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Standard XII

Physics

Electric Potential Due to Shell

A ring of rad...

Question

A ring of radius R is having two charges q and 2q distributed on its two half parts. The electric potential at a point on its axis at a distance of $2 \sqrt{2} R$ from its centre is ( $k = \frac{1}{4 π ε_{0}}$ )

\frac{3 k q}{R}

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\frac{k q}{3 R}

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\frac{k q}{R}

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\frac{k q}{\sqrt{3} R}

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Solution

The correct option is B $\frac{k q}{R}$
D= $\sqrt{R^{2} + 8 R^{2}} = 3 R$

Potential at the given point = $\frac{k q}{d} + \frac{2 k q}{d} = \frac{k q}{r}$

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Similar questions

Q. A ring of radius

R

is having two charges

q

and

2 q

distributed on its two half parts as shown in the figure below. Find the electric potential at a point on its axis at a distance

2 \sqrt{2} R

from its center.

[k = \frac{1}{4 π ε_{0}}]

Q. A ring of radius

R

is charged uniformly with a charge

+ Q

. The electric field at any point on its axis at a distance

r

from the circumference of the ring will be

Q. A ring of radius R consists two equal half parts of different charge densities. One of the half parts has charge q and other has charge 2q. The electric potential at a point on its axis at a distance of

2 \sqrt{2}

R from its centre is:

(Assume the charge density on each half part be uniform)

k = \frac{1}{4 π ε_{0}}

Consider two concentric spherical shells of radius R and 2R; with same surface charge density. If total charge on them is Q, then find electric potential at their centre.

Q. The electric filed due to a uniformly charged spherical shell of charge 'Q' and radius 'R' at a distance 'r' is for

r < R

, and for

r > R

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A ring of radius R is having two charges q and 2q distributed on its two half parts. The electric potential at a point on its axis at a distance of 2√2R from its centre is (k=14πε0)

The correct option is B kqRD=√R2+8R2=3RPotential at the given point = kqd+2kqd=kqr

A ring of radius R is having two charges q and 2q distributed on its two half parts. The electric potential at a point on its axis at a distance of $2 \sqrt{2} R$ from its centre is ( $k = \frac{1}{4 π ε_{0}}$ )

The correct option is B $\frac{k q}{R}$
D= $\sqrt{R^{2} + 8 R^{2}} = 3 R$

Potential at the given point = $\frac{k q}{d} + \frac{2 k q}{d} = \frac{k q}{r}$