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10 questions from this quiz
E=FQE = \frac{F}{Q}E=QF
NC−1\text{NC}^{-1}NC−1 and Vm−1\text{Vm}^{-1}Vm−1
E=Q4πε0r2E = \frac{Q}{4\pi\varepsilon_0 r^2}E=4πε0r2Q
So its own field doesn't distort measured field
E=VdE = \frac{V}{d}E=dV
Decreases with square of distance
Direction of force on positive test charge
Field strength magnitude and direction
8.85×10−12 F m−18.85 \times 10^{-12} \text{ F m}^{-1}8.85×10−12 F m−1
Vector quantity
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