Two charges 14 μC and -4 μC are placed at (-12 cm, 0, 0) and (12 cm, 0, 0) in an external electric field E=Br2 where B=1.2×106N/cm2 are r is in metres. The electrostatic potential energy of the configuration is

UE=kq1q2rUE = electric potential energyk =Coulomb constantq1, q2 = chargesr = distance of separationE=Br2By substituting the given values, electrostatic potential energy of the configuration is 97.9J

Two charges 14 μC and -4 μC are placed at (-12 cm, 0, 0) and (12 cm, 0, 0) in an external electric field E=Br2 where B=1.2×106N/cm2 are r is in metres. The electrostatic potential energy of the configuration is

UE=kq1q2rUE = electric potential energyk =Coulomb constantq1, q2 = chargesr = distance of separationE=Br2By substituting the given values, electrostatic potential energy of the configuration is 97.9J

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Two charges 14 μC and -4 μC are placed at (-12 cm, 0, 0) and (12 cm, 0, 0) in an external electric field $E = \frac{B}{r^{2}}$ where $B = 1 .2 \times 10^{6} N / {cm}^{2}$ are r is in metres. The electrostatic potential energy of the configuration is

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102.1 J

97.9 J

2.1 J

-97.9 J

answer is A.

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Detailed Solution

$U_{E} = k \frac{q_{1} q_{2}}{r}$
UE = electric potential energy
k =Coulomb constant
q1, q2 = charges
r = distance of separation
$E = \frac{B}{r^{2}}$
By substituting the given values, electrostatic potential energy of the configuration is 97.9J