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in CBSE by (20 points)
Two tiny spheres carrying charges \( 1.8 \mu C \) and 2.8 \( \mu C \) are located at \( 40 cm \) apart. The potential at the mid-point of the line joining the two charges is (a) \( 3.8 \times 10^{4} V \) (b) \( 2.1 \times 10^{5} V \) (c) \( 4.3 \times 10^{4} V \) (d) \( 3.6 \times 10^{5} V \)

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1 Answer

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Answer is : (B) 2.1 x 105V

Here q1 = 1.8μc = 1.8 x 10-6 C, q2 = 2.8μC = 2.8 x 10-6 C

Distance between the two spheres = 40 cm = 0.4 m

For the mid-point r1 = r2 = 0.40/2 = 0.2 m

Potential at O

V = V1+V2 \(=\frac1{4\pi\varepsilon_0}\left[\frac{q_1}{r_1}+\frac{q_2}{r_2}\right]=\frac{9\times10^9(1.8+2.8)\times10^{-6}}{0.2}\) = 2.1 x 105V

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