Question 1

An oil drop of radius 1 ~m is held stationary under a constant electric field of 3.65 10^4 ~N / C due to some excess electrons presence on it. If the density of oil drop is 1.26 ~g / cm ^3, then numbe

Accepted Answer

E=3.65 10^4 ~N / C , r=1 ~m =10^ -6 ~m aligned _ oil & =1.26 ~g / cm ^3 \ & =1.26 10^3 ~kg / m ^3 aligned Since, droplet is stationary hence weight of droplet = force due to electric field 4 3 r^3 _ oil g=q E...(i) If n be the number of excess electrons in the oil drop, then q=n e Hence, from Eq. (i), 4 3 r^3 _ oil g=n e E aligned n & = 4 r^3 _ oill g 3 e E \ & = 4 3.14 (10^ -6 )^3 1.26 10^3 10 3 1.6 10^ -19 3.65 10^4 \ & =9.03 9 aligned

Question 2

The potential of a large liquid drop when eight liquid drops are combined is 20 V , then the potential of each single drop was

Accepted Answer

Volume of eight drops = Volume of a big drop ( 4 3 r^3 ) 8= 4 3 R^3 2 r=R...(i) According to charge conservation, 8 q=Q...(ii) Potential of one small drop, V^ = q 4 _0 r Similarly, potential of big drop, V= Q 4 _0 R Now, V^ V = q Q R r V^ 20 = q 8 q 2 r r [from Eqs. (i) and (ii)] V^ =5 ~V

Question 3

A dust particle of mass 4 10^ -12 mg in suspended in air under the influence of an electric field of 50 ~N / C directed vertically upwards. How many electrons were removed from the neutral dust partic

Accepted Answer

Mass of dust particle, m=4 10^ -12 mg aligned & =4 10^ -12 10^ -3 ~g \ & =4 10^ -12 10^ -3 10^ -3 ~kg \ & =4 10^ -18 ~kg aligned Electric field, E=50 ~N / C Weight of dust particle, W=m g =4 10^ -18 10=4 10^ -17 ~N Electric force experienced by dust particle, aligned & F_e=q E \ & F_e=n e E=n 1.6 10^ -19 50 aligned where, n is the number of electrons removed from neutral dust particle. At balance condition, aligned & Electric force = Weight of dust particle \ & n 1.6 10^ -19 50=4 10^ -17 \

Question 4

Electric field at point (30,30,0) due to a charge of 0.008 C placed at origin will be, (coordinates are in cm )

Accepted Answer

E= K q r^2 r = K q r^3 r Here, aligned r & = (x_2-x_1 )^2+ (y_2-y_1 )^2+ (z_2-z_1 )^2 \ & = (30)^2+(30)^2+0^2 \ & =30 2 ~cm \ & =30 2 10^ -2 ~m aligned and q=8 10^ -3 10^ -6 C Also, r =(30 i +30 j ) 10^ -2 ~m aligned So, E & = 9 10^9 8 10^ -3 10^ -6 (30 2 10^ -2 )^3 (30 i +30 j ) 10^ -2 \ & = 9 8 10^9 10^ -11 27 2 2 10^ -6 10^3 30( i + j ) \ & = 9 8 10^2 3 27 2 2 ( i + j ) \ & =200 2 ( i + j ) NC ^ -1 aligned

Question 5

If two charges q_1 and q_2 are separated with distance ' d ' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic for

Accepted Answer

As. F(air)=F(medium) 1 4 _0 q _1 q _2 ~d _ air ^2 = 1 ~K 1 4 _0 q _1 q _2 ~d ^2 ~d _ air = d K

Question 6

Electric potential at a point 'P' due to a point charge of 5 10^ -9 C is 50 V . The distance of ' P ' from the point charge is: (Assume, 1 4 _0 =9 10^ +9 Nm ^2 C ^ -2 )

Accepted Answer

Electric potential at a point P due to a point charge, ( K =9 10^9 ) V_P= K Q r 50= 9 10^9 5 10^ -9 r r = 45 50 = 9 10 =0.9 ~m =90 ~cm

Question 7

Five charges +q,+5 q,-2 q,+3 q and -4 q are situated as shown in the figure, The electric flux due to this configuration through the surface S is :

Accepted Answer

Using Gauss's law, = q _0 Here, q = charge inside the closed surface = q+(-2 q)+5 q _0 = 4 q _0

Question 8

In a cuboid of dimension 2 ~L 2 ~L L , a charge q is placed at the centre of the surface ' S ' having area of 4 ~L ^2. The flux through the opposite surface to ' S ' is given by

Accepted Answer

After placing similar cubic at the bottom, we get cube of side 2 ~L having change q at its centre. From the Gauss's law electric flux through whole Flux passing through shaded face = q / _0 6 = q 6 _0

Question 9

The force between two small charged spheres having charges of 1 10^ -7 C and 2 10^ -7 C placed 20 ~cm apart in air is

Accepted Answer

Here, q_1=1 10^ -7 C , q_2 and 2 10^ -7 C , r=20 ~cm =20 10^ -2 ~m F= q_1 q_2 4 _0 r^2 = 9 10^9 1 10^ -7 2 10^ -7 (20 10^ -2 )^2 =4.5 10^ -3 N

Question 10

Which one of the following graphs represents the variation of electric potential with distance r from the centre of a non-conducting charged sphere of radius R ?

Accepted Answer

Electric potential at centre R=0 V_ centre =1.5 V_ surface where, aligned V_0 & = k Q R \ V_ inside & = k Q 2 R (3- r^2 R^2 ) aligned

Question 11

The electric potential V is given as a function of distance ( metre ) by V= (5 x^ 2 +10 x-4 ) volt. Question: Value of electric field at x=1 ~m is

Accepted Answer

V =5 x^ 2 +10 x-4 E = - dV dx =-(10 x+10) At x=1 m, E =-20 ~V / m

Question 12

A charge Q is placed at each of the opposite corners of a square. A charge q is placed at each of the other two corners. If the net electrical force on Q is zero, then Q q equals:

Accepted Answer

It is given that F _ A + F _ B + F _ D =0 Where F _ A , F _ B and F _ D are the forces applied by charges placed at A, B and D on the charge placed at C. array l F _ B + F _ D =- F _ A \ | F _ B + F _ D |= 2 KqQ a ^ 2 \ | F _ A |= KQ ^ 2 2 a^ 2 \ 2 KqQ a^ 2 =- KQ ^ 2 2 a^ 2 \ Q q =-2 2 array

Electrostatics — Previous Year Questions & Solutions