Two positive charges of 20 , C and Q , C are | vedclass

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(D) At the neutral point,the electric field due to both charges must be equal in magnitude and opposite in direction.Let the distance of the neutral p

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) At the neutral point,the electric field due to both charges must be equal in magnitude and opposite in direction.Let the distance of the neutral point from the $20 \, C$ charge be $r_1 = 20 \, cm = 0.2 \, m$.The total distance is $60 \, cm = 0.6 \, m$.Therefore,the distance of the neutral point from charge $Q$ is $r_2 = 60 \, cm - 20 \, cm = 40 \, cm = 0.4 \, m$.Equating the electric field magnitudes:$E_1 = E_2$$\frac{k \cdot 20}{r_1^2} = \frac{k \cdot Q}{r_2^2}$$\frac{20}{(0.2)^2} = \frac{Q}{(0.4)^2}$$\frac{20}{0.04} = \frac{Q}{0.16}$$Q = 20 	imes \frac{0.16}{0.04}$$Q = 20 	imes 4 = 80 \, C$.

Two positive charges of $20 \, C$ and $Q \, C$ are situated at a distance of $60 \, cm$. The neutral point between them is at a distance of $20 \, cm$ from the $20 \, C$ charge. The value of charge $Q$ is: (in $, C$)

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