Electrostatics Physics Solution

Physics Solution 1.

CPT_2

Two charges each equal to q( 1  3 ) are placed at the corners of an equilateral triangle of side a. The

electric field at the third corner is (a) Sol.

E3  E0

(c) E1 

(b)

where (E0  q / 4 0a2 ) (c) E3

  q q , E2  . Therefore E  E1  E 2 2 2 4 0a 4 0a 3 q

4 0 a 2

Since  1  3 , 1  3, 3  1. 

E3

E3  E0

 E12  E 22  2E1 E 2 cos 60 o 

2.

Date: 27-4-2014

3q

4 0 a

2



q

4 0 a

2



 . 4 0 a  2

Four charges are arranged at the corners of a square ABCD , as shown in the adjoining figure. The force on the charge kept at the centre O is

– 2q

B +2q

O

D

3.

Sol.

E3  E0

q

A +q

Sol.

(d)

.

 E 3  E 0  E 0  

 E0

+q C

(a) Zero (b) Along the diagonal AC (c) Along the diagonal BD (d) Perpendicular to side AB (c) We put a unit positive charge at O. Resultant force due to the charge placed at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD. Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases 4.5 times in comparison with the initial value. The ratio of the initial charges of the balls is (a) 2 (b) 3 (c) 4 (d) 6 (a) Suppose the balls having charges Q1 and Q2 respectively. Initially : Q1

Q2

A

B r

Finally : Q  Q 1 2

Q1  Q2 2

2 A

B r

 Q  Q2  k 1  2 2   k(Q1  Q2 ) F'   2 2 r r    2 2

It is given that F'  4.5F so

 (Q1  Q2 )2  4.5 Q1Q2 . On 4.

k (Q1  Q2 )2 QQ  4.5 k. 1 2 2 r2 r Q1 2  . solving it gives Q2 1

2C and 6C two charges are repelling each other with a force of 12 N . If each charge is given 2C of

charge, then the value of the force will be (b) 4 N (Repulsive) (a) 4 N (Attractive)

(c) 8 N (Repulsive)

(d) Zero Page # 1

( 2  2)  0

( 2  6)   4C

 F

Sol.

(d) Resultant charges after adding the – 2C be

5.

Three equal charges are placed on the three corners of a square. If the force between



k Q1Q2 04  k 2  0 r2 r

and that between

q1 and q 3

(a) 1 / 2 sol. 6.

Sol.

is

F13 ,

the ratio of magnitudes

(b) 2

(b) F12  1  q2 and F13  1  q 4 0 a

2

2

4 0 (a 2 )2



and

F12 F13

q1

and

q 2 is F12

is

(c) 1 / 2

(d)

2

F12 2 F13

An electron falls through a small distance in a uniform electric field of magnitude 2  10 4 NC 1 . The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be (a) Same in both cases (b) More in the case of an electron (c) More in the case of proton (d) Independent of charge 1  qE  2dm (c) The time required to fall through distance d is d    t 2 or t  2 m 

qE

Since t  m, a proton takes more time. Two similar spheres having  q and  q charge are kept at a certain distance. F force acts between the two. If in the middle of two spheres, another similar sphere having  q charge is kept, then it experience a force in magnitude and direction as (a) Zero having no direction (b) 8 F towards q charge (d) (c) 8 F towards  q charge 4 F towards  q charge 2

7.

Sol.

F k

(c) Initially, force between A and C

–Q

+Q

+Q A

r/2

B

FA FC r/2

Q2 r2

C

r

When a similar sphere B having charge +Q is kept at the mid point of line joining A and C, then Net force on B is

8.

Sol.

Fnet  FA  FC  k

r 2 Q2

2



r 2

kQ 2 2

8

kQ 2 r2

 8F

.

(Direction is shown in figure) Electric charges of 1C,  1C and 2C are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is (a) 0.9 N (b) 1.8 N (b) FA = force on C due to charge placed at A  9  10 9 

10 6  2  10 6 (10  10  2 ) 2

(c) 2.7 N

(d) 3.6 N

 1.8 N

FB = force on C due to charge placed at B  9  10 9 

10 6  2  10 6 (0.1) 2

 1.8 N

+2C C

FA 120o FB

+1C A

10 cm

– 1C B

Page # 2

Net force on C

Fnet  (FA ) 2  (FB ) 2  2FA FB cos 120 o  1.8 N

9.

Sol.

10.

Two particle of equal mass m and charge q are placed at a distance of 16 cm. They do not experience any force. The value of

q m

(a) l

(b)

 0

(c)

G

(d) They will not experience any force if | FG || Fe |

G



m2 1 q2 .  (16  10 2 )2 4 0 (16  10 2 )2

G 4 0

4 0G

(d)

q  4 0G m

Three charges each of magnitude q are placed at the corners of an equilateral triangle, the electrostatic force on the charge placed at the center is (each side of triangle is L) (a) Zero

sol.

is

q2 4 0 L2 1

(b)

(c)

3q 2 4 0 L2 1

(d)

1 q2 12 0 L2

(a) In the following figure since | FA || FB || FC | and they are equally inclined with each other, so their resultant will be zero. q A

FB

Q F C FA

q

q B

C

11.

Two point charges placed at a certain distance r in air exert a force F on each other. Then the distance r' at which these charges will exert the same force in a medium of dielectric constant k is given by (a) r

sol. 12.

sol.

13.

(c) F  F ' or

(b) r/k

Q1Q2 Q1Q2 r   r'  4 0r 2 4 0r ' 2 K K

(c) r / k

(d) r k

Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4) and (4, 1) show electrostatic attraction, while pair (2, 3) and (4, 5) show repulsion. Therefore ball 1 must be (a) Positively charged (b) Negatively charged (c) Neutral (d) Made of metal (c) Let us consider 1 ball has any type of charge. 1 and 2 must have different charges, 2 and 4 must have different charges i.e. 1 and 4 must have same charges but electrostatics attraction is also present in (1, 4) which is impossible. Equal charges q are placed at the four corners A, B, C, D of a square of length a . The magnitude of the force on the charge at B will be (a)

SOL. (c)

3q 2

4 0 a 2

4q 2

4 0 a 2

(b)

 1  2 2  q2  (c)   4 a 2 2 0  

 1  (d)  2   

q2

2  4 0 a

2

After following the guidelines mentioned above FC FD +Q A

FAC

+Q B

FA

Page # 3 D

+Q

C

Fnet  FAC  FD  FA2  FC2  FD

Since FA  FC 

(a 2 )2

1 2 2     2   

A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium, if q is equal to (a) 

Sol.

kq 2

1 2kq 2 kq 2 kq 2  q2  2  2  2   2 2  4 0a 2 2a a a 

Fnet 

14.

and FD 

kq 2 a2

(b) 

Q 2

(c) 

Q 4

Q 4

(d) 

Q 2

(b) Suppose in the following figure, equilibrium of charge B is considered. Hence for it's equilibrium | FA || FC |



1 qQ Q2  2 4 0 x 2 4 0 4 x 1

q

QA = Q A

 q

FC

C

x1

Q 4

QB = Q FA B

x2

x

Short Trick : For such type of problem the magnitude of middle charge can be determined if either of the extreme charge is in equilibrium by using the following formula. If charge A is in equilibrium then q = – QB 

x1    x 

x If charge B is in equilibrium then q  QA  2   x 

15.

2

2

If the whole system is in equilibrium then use either of the above formula. ABC is an equilateral triangle. Charges  q are placed at each corner. The electric intensity at O will be (a)

1 q 4 0 r 2

(b)

1 q 4 0 r

+q A

r

(c) Zero (d)

1

r

3q

4 0 r 2

O

+q

r +q C

B

sol. (c) EC =E EC

EB EA



120o 120o

EBC = E

EB =E

120o EA = E

 EA = E

Enet = 0

16.

The magnitude of electric field intensity E is such that, an electron placed in it would experience an electrical force equal to its weight is given by Page # 4

(b) mg

(a) mge

(c)

e

mg SOL. (b) According to the question, eE  mg  E 

e mg

(d)

e2 m2

g

e

17.

sol. 18.

An uncharged sphere of metal is placed in between two charged plates as shown. The lines of force look like ++ ++ ++ +

++ ++ ++ +

– – – – – – – A

– – – – – – – B

++ ++ ++ +

++ ++ ++ +

– – – – – – – C

– – – – – – – D

(a) A (b) B (c) C (d) D (c) Electric lines of force never intersect the conductor. They are perpendicular and slightly curved near the surface of conductor. The intensity of electric field required to balance a proton of mass 1.7  10 27 kg and charge 1.6  10 19 C is nearly (a) 1  10 7 V / m

(b) 1  10 5 V / m

mg 1.7  1027  9.8  q 1.6  1019

SOL. (a) Since qE = mg or E  19.

sol.

(c) 1  107 V / m

(d) 1  10 5 V / m

= 10.0  10–8 = 1  10–7 V/m Two point charges Q and – 3Q are placed at some distance apart. If the electric field at the location of Q is E then at the locality of 3Q , it is (a)  E (b) E / 3 (c) 3 E (d)  E / 3 (b) The field produced by charge – 3Q at A, this is E as mentioned in the Example. E

3Q x2

(along AB directed towards negative charge) A

B

Q

–3Q

x

Now field at location of – 3Q i.e. field at B due to charge Q will be E'  20.

SOL. 21.

(along AB directed

away from positive charge) The distance between a proton and electron both having a charge 1.6  10 19 coulomb , of a hydrogen atom is 10 10 metre . The value of intensity of electric field produced on electron due to proton will be (a) 2.304  10 10 N / C (b) 14.4 V / m (c) 16 V / m (d) 1.44  1011 N / C (d) E 

q

4 0 r 2

 9  109 

1.6  1019  1.44  1011 N / C (1010 )2

Two positive charges of 20 coulomb and Q coulomb are situated at a distance of 60 cm . The neutral point between them is at a distance of 20 cm from the 20 coulomb charge. Charge Q is (a) 30 C

sol.

Q E  2 3 x

(b) 40 C

(c) 60 C

(d) 80 C

(d) At neutral point Page # 5

k

22.

 Q = 80 C

20 Q k (20  10  2 )2 (40  10  2 )2

Two equal negative charge – q are fixed at the fixed points (0, a) and (0,  a) on the Y-axis. A positive charge Q is released from rest at the point (2a, 0) on the X-axis. The charge Q will (a) Execute simple harmonic motion about the origin (b) Move to the origin and remain at rest (c) Move to infinity (d) Execute oscillatory but not simple harmonic motion

sol

(d) By symmetry of problem the components of force on Q due to charges at A and B along y-axis will cancel each other while along x-axis will add up and will be along CO. Under the action of this force charge Q will move towards O. If at any time charge Q is at a distance x from O. Net force on charge Q A

–q

a



O

F Q C

x F

a

2a –q

B

Fnet  2F cos   2

i.e., Fnet  

23.

qQ x  2 2 4 0 (a  x ) (a  x 2 )1 1

2





2qQx 1 . 4 0 a 2  x 2 3

2

2

As the restoring force Fnet is not linear, motion will be oscillatory (with amplitude 2a) but not simple harmonic. A positively charged ball hangs from a silk thread. We put a positive test charge q 0 at a point and measure F / q0 , then it can be predicted that the electric field strength E

(a)  F / q0

(c)  F / q0

(b)  F / q0

(d) Cannot be estimated

Sol.

(a) Because of the presence of positive test charge q0 in front of positively charged ball, charge on the ball will be redistributed, less charge on the front half surface and more charge on the back half surface. As a result of this net force F between ball and point charge will decrease i.e. actual electric field will be greater than F / q0 .

24.

In the given figure two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of equal length L. Assume that  is so small that tan  sin  , then for equilibrium x is equal to  q2L  3    2 mg  0  

 qL2  3    2 mg  0  

1

(a) sol.

1

(b)

 q 2 L2  3    4 mg  0   1

(c)

 q2L  3    4 mg  0   1

(d)

(a)  T

T sin



T cos Fe

x mg

In equilibrium Fe = T sin ....... (i) Page # 6

mg = T cos

....... (ii)

tan  

Fe q  mg 4 o x 2  mg

Hence

x q2  2L 4 o x 2  mg

2

2  x 3  2q L

also tan   sin  





  x    4 o mg  2 o mg 

25.

(c) x 

(b) x  d

d

2

2

d 2 2

(d) x 

d 2 3

(c) Suppose third charge is similar to Q and it is q F So net force on it F Fnet = 2F cos   x  d /4 2

2



q x  d /4 2

x

Q B

Where F   Fnet  2  

d

d

2

2

1 Qq . 4 0  2 d 2  x    4  

i.e.

x x2 

d2 4

2Qqx

 d 2  4 0  x 2  4  

 d2  or  x 2  

i.e. x  

Q C

and cos 

3/2

dFnet 0 dx

     2Qqx d  0  3/2  2 dx    2 d      4 0  x  4    

 

2

x Qq 1 .  1/ 2 2 4 0  2 d 2   x    x 2  d    4   4   

For Fnet to be maximum

26.

1/ 3

Two equal charges are separated by a distance d. A third charge placed on a perpendicular bisector at x distance will experience maximum coulomb force when (a) x 

Sol.

q2L

x/2 L

3 / 2

4 

 d 2   3 x 2  x 2  4  

5 / 2 

0  

d 2 2

A small sphere carrying a charge ‘q’ is hanging in between two parallel plates by a string of length L. Time period of pendulum is T0 . When parallel plates are charged, the time period changes to T . The ratio T / T0 is equal to

Page # 7

qE   g  m    g     

   g    qE    g m  

1/ 2

(a)

Sol.

(b)

3/2

   g    qE    g m  

1/ 2

(c)

(d) None of these

(c) +

E

QE mg

–

Net downward force mg '  mg  QE

 Effect acceleration g'   g  

Hence time period T  2

27.

l l  2 QE  g'   g  m  

Three charges q1 ,  q 2 and q3 are placed as shown in the figure. The x-component of the force on q1 is proportional to (a) (b) (c) (d)

sol.

QE   m 

q2 b2 q2 b2 q2 b2 q2 b2

q3 sin  a2 Y – q3 q3  2 cos  a  q a  32 sin  a q – q1  32 cos  a 

b +q2

X

(c) – q3 a



– q1

b F2

F3 sin 



F3 cos 

+q2

F3

F2 = Force applied by q 2 on q1

F3 = Force applied by (q 3 ) on – q1

x-component of Net force on q1 is Fx = F2 + F3 sin  k

 Fx  k 

q1q2 qq  k. 1 2 3 sin  2 b a

q1q2 q1q3   2 sin   2 a  b 

 Fx  k  q1 

q2

b

28.

2



q3 q  q  sin    Fx   22  32 sin   a2 a  b 

Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U respectively Q P (a) , , , , ,  Page # 8 U

R O

(b) , , , , ,  (c) , , , , ,  (d) , , , , ,  sol.

(d) If the charges are arranged according to the option (d), the electric fields due to P and S and due to Q and T add to zero, while due to U and R will be added up.

29.

Two identical point charges are placed at a separation of d. P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E, E is plotted against x for values of x from close to zero to slightly less than d. Which of the following represents the resulting curve

(a)

Y

(b)

Y

E

E

O

(c)

Y

X

x

(d)

E x

Sol.

30.

Sol.

X

x

Y

E O

O

X

x O

X

(d) At mid point, E = 0 Before mid point, E is positive. This is maximum near the charge and decreases towards mid point. After mid point, E is negative, The curve crosses x-axes at x-axis at x = d/2. From centre to end, E decreases. The variation is shown by curve. Assertion : Charge is quantized Reason : Charge, which is less than 1 C is not possible (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If assertion is false but reason is true. (c) Q  ne and charge lesser than 1 C is possible.

Page # 9

Page # 10