Two equal point charges are fixed at x = -α and x = +α on the X-axis . Another point charge Q is placed at the origin. The change in electrical potential energy of Q when it is displaced by a small distance x along the X-axis, is approximately proportional to
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a
X
b
X2
c
X3
d
1/X
answer is B.
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Detailed Solution
When Q is at position x, thenU=qQ4πε01(a+x)+1(a−x)or U=2qQa4πε0a2−x2∴ dUdx=2aqQ4πε02xa2−x22 =2aqQ4πε0×2xa4 ∵a2−x2≈a2Now dU∝dx or U∝x2.