A charge of 0.8 text{ C} is divided into two | vedclass

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(A) Let the total charge be $Q = 0.8 	ext{ C}$.Let $Q_{1} = q$,then $Q_{2} = Q - q = (0.8 - q)$.The electrostatic force between the two charges is gi

Vedclass · Accepted Answer

$Q_{1} = Q_{2} = 0.4 	ext{ C}$

Vedclass · Answer

(A) Let the total charge be $Q = 0.8 	ext{ C}$.Let $Q_{1} = q$,then $Q_{2} = Q - q = (0.8 - q)$.The electrostatic force between the two charges is given by Coulomb's Law:$F = \frac{k Q_{1} Q_{2}}{r^{2}} = \frac{k q (0.8 - q)}{r^{2}}$For the force $F$ to be maximum,the derivative of $F$ with respect to $q$ must be zero:$\frac{dF}{dq} = \frac{k}{r^{2}} \frac{d}{dq} (0.8q - q^{2}) = 0$$0.8 - 2q = 0$$2q = 0.8$$q = 0.4 	ext{ C}$Thus,$Q_{1} = 0.4 	ext{ C}$ and $Q_{2} = 0.8 - 0.4 = 0.4 	ext{ C}$.Therefore,the force is maximum when $Q_{1} = Q_{2} = 0.4 	ext{ C}$.

$A$ charge of $0.8 \text{ C}$ is divided into two charges $Q_{1}$ and $Q_{2}$. These are kept at a separation of $30 \text{ cm}$. The force on $Q_{1}$ is maximum when

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