Question

four charges of + 6 into 10 raise to minus 8 coulomb each and placed at the corners of square whose sides are 3 cm each calculate the resultant force on a charge and show that I direction and shows in direction and a diagram drawn to scale

Answer 1

The resultant force on a charge is 0.0689 N

Direction: North of west

Explanation:

Four charges of $+6\times 10^{-8}$ coulomb each and placed at the corners of square whose sides are 3 cm

Al changes are identical.

Let $+6\times 10^{-8}=q$

Distance between them , r = 3 cm = 0.03 m

Please see attachment for figure.

Force on A through B, $F_{BA}$

$F_{BA}=\dfrac{kq^2}{r^2}$

Force on A through D, $F_{DA}$

$F_{DA}=\dfrac{kq^2}{r^2}$

Force on A through C, $F_{CA}$

$F_{CA}=\dfrac{kq^2}{2r^2}$

Let the resultant force on A through system of charges, F

$F=F_{CA}+\sqrt{F_{BA}^2+F_{DA}^2}$

$F=\dfrac{kq^2}{2r^2}+\dfrac{kq^2}{r^2}\sqrt{1+1}$

$F=\dfrac{kq^2}{r^2}(\dfrac{1}{2}+\sqrt{2})$

where, $k=9\times 10^9, q=6\times 10^{-8},r=3\times 10^{-2}$

$F=\dfrac{9\times 10^9\times (6\times 10^{-8})^2}{(3\times 10^{-2})^2}(\dfrac{1}{2}+\sqrt{2})$

$F=0.0689\ N$

Hence, the resultant force on a charge is 0.0689 N

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