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1. Electric Charges and Fields
normal
Calculate the work done in taking a charge $-2 \times 10^{-9} \,C$ from $A$ to $B$ via $C$ is ......... (in diagram)

A
$0.2$
B
$1.2$
C
$2.2$
D
$0$
Solution

(b)
$V_A=\frac{9 \times 10^9 \times 8 \times 10^{-3}}{3 \times 10^{-2}}=24 \times 10^8 \,V$
$V_B=\frac{9 \times 10^9 \times 8 \times 10^{-3}}{4 \times 10^{-2}}=18 \times 10^8 \,V$
$\Delta W=-2 \times 10^{-9}\left(-6 \times 10^8\right)$
$\Delta W=12 \times 10^{-1} \,J$
$\Delta W=1.2\,J$
Standard 12
Physics