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Question

Physics Question on coulombs law

Two identical metal spheres charged with +12μF+12 \mu F and 8μF-8 \mu F are kept at certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is

A

12:01

B

8:01

C

24:01:00

D

4:01

Answer

24:01:00

Explanation

Solution

Finitial =14πε0×12×(8)r2;Finitial =14πε0×96r2F_{\text {initial }}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{12 \times(-8)}{r^{2}} ; F_{\text {initial }}=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{96}{r^{2}}
where rr is the distance between them. When the charges are brought in contact, then
q1=q2=1282=42=2μFq_{1} =q_{2}=\frac{12-8}{2}=\frac{4}{2}=2 \mu F
Ffinal =14πε0×2×2r2=4r2×14πε0\therefore \,\,\,\,F_{\text {final }} =\frac{1}{4 \pi \varepsilon_{0}} \times \frac{2 \times 2}{r^{2}}=\frac{4}{r^{2}} \times \frac{1}{4 \pi \varepsilon_{0}}
Ffinal =4r2×14πε0\Rightarrow |F|_{\text {final }}=\frac{4}{r^{2}} \times \frac{1}{4 \pi \varepsilon_{0}}
Finitial Ffinal =964=24\therefore \frac{|F|_{\text {initial }}}{|F|_{\text {final }}}=\frac{96}{4}=24