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Potential Energy - External Viewpoint

In given Figu...

Question

In given Figure, a charge Q is fixed. Another charge q is moved along a circular arc MN of radius r around it, from the point M to the point N such that the length of the arc $M N = l$ . The work done in this process is :

Zero

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\frac{1}{4 π ϵ_{0}} . \frac{Q q}{r^{2}} l

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\frac{Q q}{2 ϵ_{0} r^{2}} l

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\frac{Q q}{2 π ϵ_{0} r^{2}}

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Solution

The correct option is A Zero
Since the point M and N are at equal distances from point Q. So these will have same potential.
$V_{M} = V_{N} \frac{1}{4 π ϵ_{0}} \frac{q}{r}$
$∴ V_{M} - V_{N} = 0$
The work done in this process
$W = (V_{M} - V_{N}) q = 0$

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In given Figure, a charge Q is fixed. Another charge q is moved along a circular arc MN of radius r around it, from the point M to the point N such that the length of the arc MN=l. The work done in this process is :

The correct option is A ZeroSince the point M and N are at equal distances from point Q. So these will have same potential.VM=VN14πϵ0qr∴VM−VN=0The work done in this processW=(VM−VN)q=0

In given Figure, a charge Q is fixed. Another charge q is moved along a circular arc MN of radius r around it, from the point M to the point N such that the length of the arc $M N = l$ . The work done in this process is :

The correct option is A Zero
Since the point M and N are at equal distances from point Q. So these will have same potential.
$V_{M} = V_{N} \frac{1}{4 π ϵ_{0}} \frac{q}{r}$
$∴ V_{M} - V_{N} = 0$
The work done in this process
$W = (V_{M} - V_{N}) q = 0$