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Question

Two points charges 4q andq are fixed on thex-axis at x=-d2 and x=d2, respectively. If a third point charge qis taken from the origin to x=d along the semicircle as shown in the figure, the energy of the charge will:


A

decrease by q24πεod

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B

decrease by 4q23πεod

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C

increases by3q24πεod

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D

decrease by 2q23πεod

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Solution

The correct option is B

decrease by 4q23πεod


Step 1: Given

  1. The value of the two-point charges 4q andq are at x=-d2 and x=d2.
  2. The value of the third charge is q at x=d.

Step 2: To find

The energy of the third charge

Step 3: Solution

The potential energy U=q1q24πεoR

q1=Valueoffirstchargeq2=ValueofsecondchargeR=Distancebetweenthecharges

Let initial and final potential energy be Uiand Uf

Potential energy at initial point = potential energy due to 4q and the moving charge q at A + potential energy due -q and the moving charge q at A.

Ui=14πεo4q×qd2+q-qd2Ui=14πεo4q2d2-q2d2

Potential energy at final point = potential energy due 4q and the moving charge q at B + potential energy due -q and the moving charge q at B.

Uf=14πεo4q×q3d2+q-qd2Uf=14πεo4q23d2-q2d2

The change in the energy

U=Uf-Ui

U=14πεo4q23d2-4q2d2U=4q24πεo23d-2dU=-4q23πεod

Here, the negative sign shows that energy is decreasing.

Hence, the correct option is B.


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