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A spherical drop of water has 3 x 10⁻¹⁰ C amount of charge residing on it. 500 V electric potential exists on its surface. Calculate the radius of this drop. If eight such drops (Having identical charge and radii) combine to form a single drop, calculate the electric potential on the surface of the new drop. k = 9 x 10⁹ SI. |
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Answer» udent,◆ Answer -r = 5.4 mmV' = 2000 V● Explanation -Let r be the radius of the drop. Electric potential of the drop is GIVEN by -V = (1/4πε) q/rr = (1/4πε) q/Vr = 9×10^9 × 3×10^-10 / 500r = 5.4×10^-3 MWHEN 8 such drops are merged together, radius of bigger drop will be -R = r.∛nR = 5.4×10^-3 × ∛8R = 5.4×10^-3 × 2R = 1.08×10^-2 mTotal CHARGE on bigger drop will be -Q = nqQ = 8 × 3×10^-10Q = 2.4×10^-9 CNow, electric potential of bigger drop will be -V' = (1/4πε) Q/RV' = 9×10^9 × 2.4×10^-9 / 1.08×10^-2V' = 2000 VHope this helps you... |
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