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Electrostatics

NEET Physics · 98 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
Four point charges −-Q, −-q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero is
A Q = −-q
B Q = −1q - {1 \over q}
C Q = q
D Q = 1q{1 \over q}
Correct Answer
Option A
Solution

Let the side length of square be 'a' then potential at centre O is

V=k(−Q)a2+k(−q)a2+k(2q)a2+k(2Q)a2=0V = {{k( - Q)} \over {{a \over {\sqrt 2 }}}} + {{k( - q)} \over {{a \over {\sqrt 2 }}}} + {{k(2q)} \over {{a \over {\sqrt 2 }}}} + {{k(2Q)} \over {{a \over {\sqrt 2 }}}} = 0

= – Q – q + 2q + 2Q = 0 = Q + q = 0 ∴\therefore Q = – q

Q42
What is the flux through a cube of side aa if a point charge of q is at one of its corner?
A 2qε0{{2q} \over {{\varepsilon _0}}}
B q8ε0{q \over {8{\varepsilon _0}}}
C qε0{q \over {{\varepsilon _0}}}
D q2ε06a2{q \over {2{\varepsilon _0}}}6{a^2}
Correct Answer
Option B
Solution

Eight identical cubes are required so that the given charge q appears at the centre of the bigger cube.

Thus, the electric flux passing through the given cube is

ϕ=18(qε0)=q8ε0\phi = {1 \over 8}\left( {{q \over {{\varepsilon _0}}}} \right) = {q \over {8{\varepsilon _0}}}
Q43
An electric dipole of moment p is placed in an electric field of intensity E. The dipole acquires a position such that the axis of the dipole makes an angle θ\theta with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ\theta = 90 o , the torque and the potential energy of the dipole will respectively be
A pEsinθ\theta , −-pEcosθ\theta
B pEsinθ\theta , −-2pEcosθ\theta
C pEsinθ\theta , 2pEcosθ\theta
D pEcosθ\theta , −-pEsinθ\theta
Correct Answer
Option A
Solution

Torque, τ\tau = pEsinθ\theta Potential energy, U = –pEcosθ\theta

Q44
Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC, 2aa. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is
A 3qQ4πε0a{{3qQ} \over {4\pi {\varepsilon _0}a}}
B 3qQ8πε0a{{3qQ} \over {8\pi {\varepsilon _0}a}}
C qQ4πε0a{{qQ} \over {4\pi {\varepsilon _0}a}}
D zero
Correct Answer
Option D
Solution

AC = BC V D = V E We have, W = Q (V E – V D ) ⇒\Rightarrow W = 0

Q45
The electric potential V at any point (x, y, z), all in metres in space is given by V = 4x 2 volt. The electric field at the point (1, 0, 2) in volt/meter, is
A 8 along negative X-axis
B 8 along positive X-axis
C 16 along negative X-axis
D 16 along positive X-axis
Correct Answer
Option A
Solution
E→=−∇→V\overrightarrow E = - \overrightarrow \nabla V

where

∇→=i^∂∂x+j^∂∂y+∂∂z\overrightarrow \nabla = \widehat i{\partial \over {\partial x}} + \widehat j{\partial \over {\partial y}} + {\partial \over {\partial z}}
E→=−[i^∂V∂x+j^∂V∂y+k^∂V∂z]\overrightarrow E = - \left[ {\widehat i{{\partial V} \over {\partial x}} + \widehat j{{\partial V} \over {\partial y}} + \widehat k{{\partial V} \over {\partial z}}} \right]

Here, V = 4x 2

∴E→=−8xi^\therefore \overrightarrow E = - 8x\widehat i

The electric field at point (1, 0, 2) is

E→(1,0,2)=−8i^Vm−1{\overrightarrow E _{\left( {1,0,2} \right)}} = - 8\widehat iV{m^{ - 1}}

So electric field is along the negative X-axis.

Q46
Four electric charges +q, +q, −- q and −- q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, midway between the two charges + q and +q, is
A 14πε02qL(1+5){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 + \sqrt 5 } \right)
B 14πε02qL(1+15){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 + {1 \over {\sqrt 5 }}} \right)
C 14πε02qL(1−15){1 \over {4\pi {\varepsilon _0}}}{{2q} \over L}\left( {1 - {1 \over {\sqrt 5 }}} \right)
D zero
Correct Answer
Option C
Solution

Distance of point A from the two +q charges = L. Distance of point A from the two –q charges

=L2+(2L)2=5L= \sqrt {{L^2} + {{\left( {2L} \right)}^2}} = \sqrt 5 L

∴\therefore

VA=(KqL×2)−(Kq5L×2){V_A} = \left( {{{Kq} \over L} \times 2} \right) - \left( {{{Kq} \over {\sqrt 5 L}} \times 2} \right)
=2KqL[1−15]= {{2Kq} \over L}\left[ {1 - {1 \over {\sqrt 5 }}} \right]
=14πε0.2qL(1−15)= {1 \over {4\pi {\varepsilon _0}}}.{{2q} \over L}\left( {1 - {1 \over {\sqrt 5 }}} \right)
Q47
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
A increase four times
B be reduced to half
C remain the same
D be doubled
Correct Answer
Option C
Solution

According to Gauss’s law

ϕE=Qenclosedε0{\phi _E} = {{{Q_{enclosed}}} \over {{\varepsilon _0}}}

If the radius of the Gaussian surface is doubled, the outward electric flux will remain the same.

This is because electric flux depends only on the charge enclosed by the surface.

Q48
The electric field at a distance 3R2{{3R} \over 2} from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R2{R \over 2} from the centre of the sphere is
A zero
B E
C E2{E \over 2}
D E3{E \over 3}
Correct Answer
Option A
Solution

Electric field inside a charged conductor is always zero.

Q49
A square surface of side L meter in the plane of the paper is placed in a uniform electric field EE(volt/m) acting along the same plane at an angle θ\theta with the horizontal side of the square as shown in figurre. The electric flux linked to the surface, in units of volt m is
A EL 2
B EL 2 cosθ\theta
C EL 2 sinθ\theta
D zero
Correct Answer
Option D
Solution

Electric flux, ϕ\phi = EA cos θ\theta , where θ\theta = angle between E and normal to the surface. Here,

θ=π2\theta = {\pi \over 2}
⇒ϕ\Rightarrow \phi

= 0

Q50
Two positives ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)
A 4πε0Fd2e2{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}
B 4πε0Fe2d2\sqrt {{{4\pi {\varepsilon _0}F{e^2}} \over {{d^2}}}}
C 4πε0Fd2e2\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}}
D 4πε0Fd2q2{{4\pi {\varepsilon _0}F{d^2}} \over {{q^2}}}
Correct Answer
Option C
Solution

According to Coulomb’s law, the force of repulsion between the two positive ions each of charge q, separated by a distance d is given by

F=14πε0(q)(q)d2F = {1 \over {4\pi {\varepsilon _0}}}{{\left( q \right)\left( q \right)} \over {{d^2}}}
F=q24πε0d2F = {{{q^2}} \over {4\pi {\varepsilon _0}{d^2}}}
q2=4πε0Fd2{q^2} = 4\pi {\varepsilon _0}F{d^2}
q=4πε0Fd2q = \sqrt {4\pi {\varepsilon _0}F{d^2}}

...(i) Since, q = ne where, n = number of electrons missing from each ion e = magnitude of charge on electron

∴n=qe\therefore n = {q \over e}
n=4πε0Fd2en = {{\sqrt {4\pi {\varepsilon _0}F{d^2}} } \over e}

(Using (i)) =

4πε0Fd2e2\sqrt {{{4\pi {\varepsilon _0}F{d^2}} \over {{e^2}}}}
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