Two charges+Q each are fixed at points C and D. Line AB is the bisector line of CD. A third charge +q is moved from A to B, then from B to C

From A to B electrostatic potential energy will decrease

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From A to B electrostatic potential energy will increase

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From B to C electrostatic potential energy will increase

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From B to C electrostatic potential energy will decrease

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Solution

The correct options are
B From A to B electrostatic potential energy will increase
C From B to C electrostatic potential energy will increase
We assume , $B C = B D = B A = r$ so, $A C = A D = \sqrt{r^{2} + r^{2}} = \sqrt{2} r$
Potential energy at A is $U_{A} = k \frac{Q q}{A C} + k \frac{Q q}{A D} = k \frac{Q q}{\sqrt{2} r} + k \frac{Q q}{\sqrt{2} r} = k \frac{2 Q q}{\sqrt{2} r}$
Potential energy at B is $U_{B} = k \frac{Q q}{B C} + k \frac{Q q}{B D} = k \frac{Q q}{r} + k \frac{Q q}{r} = k \frac{2 Q q}{r}$
Potential energy at C is $U_{C} = k \frac{Q q}{0} + k \frac{Q q}{B D} \sim \infty .$
Thus, $U_{A} < U_{B} < U_{C}$

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