Practicing Success
Practicing Success
CUET Preparation Today
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$\text{Potential due to charge on Sphere }V_1= \frac{q}{4\pi\epsilon_0R}$ $\text{Potential due to charge Q placed at Sphere } V_2= \frac{Q}{4\pi\epsilon_0R/2} = \frac{2Q}{4\pi\epsilon_0R} $ $\text{Net potential } V = V_1 + V_2 = \frac{2Q}{4\pi\epsilon_0R} + \frac{q}{4\pi\epsilon_0R}$
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