Question

Two point charges each of magnitude 20mc separated by a distance of 1.5m in a large vessel of water. The relative permitivity of water is 80. Calculate the force between them

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Force=0.02N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt:\implies Magnitude \: of \: point \: charges = 20 \: \mu C\\ \\ \tt: \implies Distance \: between \: them(r) = 1.5 \: m \\ \\ \tt: \implies Relative \: permitivity( \epsilon_{r}) = 80 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Force \: between \: them(F) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F= \frac{1}{4\pi \: \epsilon} \frac{q_{1} q_{2} }{ {r}^{2} } \\ \\ \tt: \implies F = \frac{1}{4\pi \: \epsilon_{o} \: \epsilon_{r} } \frac{20 \times 20}{ {1.5}^{2} } \\ \\ \tt: \implies F= \frac{9 \times {10}^{ 9} \times 400 \times {10}^{ - 12} }{80 \times 2.25} \\\\ \tt:\implies F=\frac{9\times 4\times 10^{9-10}}{180} \\\\ \tt: \implies F= \frac{3.6}{180} \\ \\ \green{\tt: \implies F= 0.02 \: N} \\ \\ \green{\tt \therefore Force \: between \: them \: is \: of \: 0.02 \: N}$

Answer 2

GIVEN :-

Magnitude of the point charges = $\sf{20\mu}$

Distance between them(r) = 1.5 m

Relative permitivity of water($\sf{\epsilon_r}$) = 80

TO FIND :-

The force between them.

FORMULA TO BE USED :-

$\displaystyle{\sf{F=\frac{1}{4\pi \epsilon}\times \frac{q_1q_2}{r^2} }}$

SOLUTION :-

Putting the known values in the above formula :-

$\displaystyle{\sf{\implies F=\frac{1}{4\pi \epsilon_o \epsilon_r}\times \frac{20^2}{1.5^2} }}\\\\\\\displaystyle{\sf{\implies F=\frac{9\times 10^9\times400\times10^{-12}}{2.25\times80} }}\\\\\\\displaystyle{\sf{\implies F=\frac{3600\times 10^{9+(-12)}}{180} }}\\\\\\\displaystyle{\sf{\implies F=20\times10^{-3} }}\\\\\boxed{\displaystyle{\sf{\implies F=0.02\ N }}}$

∴ Thus, the force between them is 0.02 N.