Electric field in a region is given as $\to E = 2^i$ . A charge of $1 C$ is placed at $x = 1 m$ and a charge of $2 C$ is placed at $x = 2 m$ . Find the total energy of the system of charges. (Assume potential at $x = 0$ is zero)

Zero

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\frac{2}{4 π ε_{o}} - 10

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\frac{2}{4 π ε_{o}} - 6

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$- \frac{2}{4 π ε_{o}} + 10$

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Solution

The correct option is B $\frac{2}{4 π ε_{o}} - 10$
$V = - \int \to E . \to d r = - 2 x$
At $x = 1, V_{1} = - 2$
At $x = 2, V_{2} = - 4$
Potential energy of system, $U = \frac{q_{1} q_{2}}{4 π ε_{o} d} + q_{1} V_{1} + q_{2} V_{2}$
Substituting values, $U = \frac{2}{4 π ε_{o}} - 10$

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Electric field in a region is given as →E=2^i. A charge of 1C is placed at x=1m and a charge of 2C is placed at x=2m. Find the total energy of the system of charges. (Assume potential at x=0 is zero)

The correct option is B 24πεo−10V=−∫→E.→dr=−2xAt x=1,V1=−2At x=2,V2=−4Potential energy of system, U=q1q24πεod+q1V1+q2V2Substituting values, U=24πεo−10

Electric field in a region is given as $\to E = 2^i$ . A charge of $1 C$ is placed at $x = 1 m$ and a charge of $2 C$ is placed at $x = 2 m$ . Find the total energy of the system of charges. (Assume potential at $x = 0$ is zero)

The correct option is B $\frac{2}{4 π ε_{o}} - 10$
$V = - \int \to E . \to d r = - 2 x$
At $x = 1, V_{1} = - 2$
At $x = 2, V_{2} = - 4$
Potential energy of system, $U = \frac{q_{1} q_{2}}{4 π ε_{o} d} + q_{1} V_{1} + q_{2} V_{2}$
Substituting values, $U = \frac{2}{4 π ε_{o}} - 10$