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(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \mathrm{C}$ ? The radii of $A$ and $B$ are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
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Verified Answer
(a) Given, $\mathrm{q}_1=6.5 \times 10^{-7} \mathrm{C}$,
$\mathrm{q}_2=6.5 \times 10^{-7} \mathrm{C}$,
$\mathrm{r}=50 \mathrm{~cm}=0.5 \mathrm{~m}, \mathrm{~F}=$ ?
By formula, $\quad F=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$
$$
=\frac{9 \times 10^9 \times 6.5 \times 10^{-7} \times 6.5 \times 10^{-7}}{(0.5)^2}
$$
$=1.521 \times 10^{-2} \mathrm{~N}$
(b) Doubling charge on each sphere increases the force four times. Making the distance half, further increases the force four times.
So new force becomes 16 times.
Now $\mathrm{F}=1.521 \times 10^{-2} \times 16$ $=2434 \times 10^{-2}=0.2434 \mathrm{~N}$
$\mathrm{q}_2=6.5 \times 10^{-7} \mathrm{C}$,
$\mathrm{r}=50 \mathrm{~cm}=0.5 \mathrm{~m}, \mathrm{~F}=$ ?
By formula, $\quad F=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$
$$
=\frac{9 \times 10^9 \times 6.5 \times 10^{-7} \times 6.5 \times 10^{-7}}{(0.5)^2}
$$
$=1.521 \times 10^{-2} \mathrm{~N}$
(b) Doubling charge on each sphere increases the force four times. Making the distance half, further increases the force four times.
So new force becomes 16 times.
Now $\mathrm{F}=1.521 \times 10^{-2} \times 16$ $=2434 \times 10^{-2}=0.2434 \mathrm{~N}$
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