Point charge q1 - - 2-40 - C is held stationary at the origin- A second point charge q2 - - 4-30 -C moves from the point x - 0-150m- y- 0 to the point x - 0-250m- y- 0-250m- The amount of work is done by the electric force on q2 is - 356 - 10 -xJ- Find x-

PE at final point = PE at initial point + work done nbsp;Work done = PE at final point PE at initial point = nbsp;1/4 πϵq_1 q_2/√(0.15^2 + 0^2) nbsp; ...

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Byju's Answer

Standard XII

Physics

Potential Energy of a System of Multiple Point Charges

point charge ...

Question

point charge $q_{1} = + 2.40 μ C$ is held stationary at the origin. A second point charge $q_{2} = - 4.30 μ$ C moves from the point $x = 0.150 m, y = 0$ to the point $x = 0.250 m, y = 0.250 m .$ The amount of work is done by the electric force on $q_{2}$ is $- 356 \times 10^{- x} J$ . Find x.

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Solution

PE at final point = PE at initial point + work done
Work done = PE at final point - PE at initial point = $\frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{\sqrt{{0.15}^{2} + 0^{2}}} - \frac{1}{4 π ϵ} \frac{q_{1} q_{2}}{\sqrt{{0.25}^{2} + {0.25}^{2}}} = 0.356 J$
Work done by electric force = - work done = -0.356 J

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point charge q1=+2.40μC is held stationary at the origin. A second point charge q2=−4.30μC moves from the point x=0.150m,y=0 to the point x=0.250m,y=0.250m. The amount of work is done by the electric force on q2 is −356×10−xJ. Find x.

PE at final point = PE at initial point + work done Work done = PE at final point - PE at initial point = 14πϵq1q2√0.152+02−14πϵq1q2√0.252+0.252=0.356JWork done by electric force = - work done = -0.356 J

point charge $q_{1} = + 2.40 μ C$ is held stationary at the origin. A second point charge $q_{2} = - 4.30 μ$ C moves from the point $x = 0.150 m, y = 0$ to the point $x = 0.250 m, y = 0.250 m .$ The amount of work is done by the electric force on $q_{2}$ is $- 356 \times 10^{- x} J$ . Find x.