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Numericals · Q18

Q.Two small spheres 18 cm apart have equal negative charges and repel each other with the force of 6×10−36\times10^{-3} N. Find the total charge on both spheres.

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Since the two spheres carry equal charges q (both negative) and repel with force F=6×10−3F=6\times10^{-3} N at separation r=0.18r=0.18 m, Coulomb's law gives F=14πϵ0q2r2  ⇒  q2=Fr2k=(6×10−3)(0.18)29×109=(6×10−3)(0.0324)9×109=1.944×10−49×109=2.16×10−14.F=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}\;\Rightarrow\;q^2=\frac{Fr^2}{k}=\frac{(6\times10^{-3})(0.18)^2}{9\times10^9}=\frac{(6\times10^{-3})(0.0324)}{9\times10^9}=\frac{1.944\times10^{-4}}{9\times10^9}=2.16\times10^{-14}. Taking the square root, q=2.16×10−14≈1.47×10−7q=\sqrt{2.16\times10^{-14}}\approx1.47\times10^{-7} C on EACH sphere. The question asks for the TOTAL charge on BOTH spheres combined, which is qtotal=2q≈2.94×10−7 C  (≈0.294 μC).q_{total}=2q\approx2.94\times10^{-7}\,\text{C}\;(\approx0.294\,\mu\text{C}). Disclosed discrepancy: the source text's printed bracketed answer reads 'q=2.94×10−10q=2.94\times10^{-10} C', three orders of magnitude smaller than this independently-verified calculation (using the given force and separation exactly as stated) -- consistent with a decimal-exponent misprint or OCR corruption of the original scanned figure, since the leading digits '2.94' match exactly; the corrected value, 2.94×10−72.94\times10^{-7} C, is presented here as the physically correct answer. [!ANSWER] Total charge on both spheres ≈2.94×10−7\approx2.94\times10^{-7} C.

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