The electric potential V at any point (x, y,z), all in meters in space is given by V = 4x² volt. The electric field at the point(1, 0, 2) in volt/meter is

8 along positive X-axis
16 along negative X-axis
16 along positive X-axis
8 along negative X-axis

Solution

A square surface of side L meter in the plane of the paper is placed in a uniform electric field E(volt/m) acting along the same plane at an angle θ with the horizontal side of the square as shown in Figure. The electric flux linked to the surface, in units of volt. m, is

EL²
EL² cosθ
EL² sinθ
zero

Solution

Electric flux, φ= EA cos θ, where θ

= angle between E and normal to the surface.

Here θ=^π⁄₂

⇒ φ = 0

Two positive 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 of an electron)

\(\frac{4\pi \varepsilon _{0}Fd^{2}}{e^{2}}\)
\(\sqrt{\frac{4\pi \varepsilon _{0}Fe^{2}}{d^{2}}}\)
\(\sqrt{\frac{4\pi \varepsilon _{0}Fd^{2}}{e^{2}}}\)
\(\frac{4\pi \varepsilon _{0}Fd^{2}}{q^{2}}\)

Solution

The electric intensity due to a dipole of length 10 cm and having a charge of 500 mC, at a point on the axis at a distance 20 cm from one of the charges in air, is

6.25× 10⁷N/C
9.28 × 10⁷N/C
13.1 × 10¹¹N/C
20.5 × 10⁷N/C

Solution

If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is

Q/2
–Q/2
Q/4
–Q/4

Solution

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

Solution

Electric lines of force never intersect the conductor.They are perpendicular and slightly curved near the surface of conductor.

A solid conducting sphere of radius a has a net positive charge 2Q. A conducting spherical shell of inner radius band outer radius c is concentric with the solid sphere and has a net charge – Q.The surface charge density on the inner and outer surfaces of the spherical shell will be

\(-\frac{2Q}{4\pi b^{2}},\frac{Q}{4\pi c^{2}}\)
\(-\frac{Q}{4\pi b^{2}},\frac{Q}{4\pi c^{2}}\)
\(0,\frac{Q}{4\pi c^{2}}\)
None of these

Solution

Electric lines of force never intersect the conductor.They are perpendicular and slightly curved near the surface of conductor.

A glass rod rubbed with silk is used to charge a gold leaf electroscope and the leaves are observed to diverge. The electroscope thus charged is exposed to X-rays for a short period. Then

the divergence of leaves will not be affected
the leaves will diverge further
the leaves will collapse
the leaves will melt

Solution

Charge on glass rod is positive, so charge on gold leaves will also be positive.Due to X-rays, more electrons from leaves will be emitted, so leaves becomes more positive and diverge further.

A solid sphereical conductor of radius R has a spherical cavity of radius a (a < R) at its centre. A charge + Q is kept at the centre. The cahrge at the inner surface, outer and at a positionr (a < r < R) are respectively

+Q, – Q, 0
– Q, + Q,0
0, – Q, 0
+Q ,0, 0

Solution

The charge at the inner surface, outer surface and inside the conductor at P= (– Q, +Q, 0) as shown in the figure

The electrostatic potential (φ_r) of a spherical symmetric system kept at origin, is shown in the figure, and given as

Which of the following option(s) is/are incorrect

a. For spherical region r ≤ R₀ total electrostatic energy stored is zero
b. Within r = 2R₀ total charge is q.
c. There will be no charge anywhere except at r =R₀
d. None of these

Solution

(a, b, c, d)
The potential shown is for charged spherical conductor.

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