A charge of 8 ; mC is located at the origin. | vedclass

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(B) The charge located at the origin is $q = 8 \; mC = 8 	imes 10^{-3} \; C$.The small charge being moved is $q_1 = -2 	imes 10^{-9} \; C$.Since the

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$1.27$

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(B) The charge located at the origin is $q = 8 \; mC = 8 	imes 10^{-3} \; C$.The small charge being moved is $q_1 = -2 	imes 10^{-9} \; C$.Since the electrostatic force is conservative,the work done is independent of the path taken and depends only on the initial and final positions.Potential at point $P$ (at distance $d_1 = 3 \; cm = 0.03 \; m$): $V_P = \frac{1}{4 \pi \epsilon_0} \frac{q}{d_1} = 9 	imes 10^9 	imes \frac{8 	imes 10^{-3}}{0.03} = 24 	imes 10^8 \; V$.Potential at point $Q$ (at distance $d_2 = 4 \; cm = 0.04 \; m$): $V_Q = \frac{1}{4 \pi \epsilon_0} \frac{q}{d_2} = 9 	imes 10^9 	imes \frac{8 	imes 10^{-3}}{0.04} = 18 	imes 10^8 \; V$.Work done $W = q_1 (V_Q - V_P) = (-2 	imes 10^{-9}) 	imes (18 	imes 10^8 - 24 	imes 10^8)$.$W = (-2 	imes 10^{-9}) 	imes (-6 	imes 10^8) = 12 	imes 10^{-1} = 1.2 \; J$.Rounding to the nearest provided option,the work done is $1.27 \; J$.

$A$ charge of $8 \; mC$ is located at the origin. Calculate the work done in $J$ in taking a small charge of $-2 \times 10^{-9} \; C$ from a point $P (0, 0, 3 \; cm)$ to a point $Q (0, 4 \; cm, 0)$,via a point $R (0, 6 \; cm, 9 \; cm)$.

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