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Asked by chingitamkarthik | 07 Feb, 2024, 09:33: AM

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As can be seen from above figure , sphere of radius 4a centered at origin encloses only charge 5Q placed at (3a, 0).

Hence flux through the surface of sphere = (5Q)/ε_o

Answered by Thiyagarajan K | 07 Feb, 2024, 01:22: PM

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Asked by chingitamkarthik | 07 Feb, 2024, 09:33: AM

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27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is that of a smaller drop. times

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point charge 1.5*10to the power -3 and 0.5 are placed at corner A and C of a right angled triangle ABC such that AB= 1.2m and BC= 0.5m find the resultant electric field at B

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