A point charge 25 μC is located in the XY-plane at the point of position vector r0→=i^+j^m. What is the magnitude of electric field at the point of position vector r1→=4i^+5j^m?

E= kqr2 wherer= r1→-r0→=4i^+5j^-i^-j^=3i^+4j^= 5 m⇒E=9 x 109 x 25 x 10-65 x 5 =9 x 103 Vm-1=9kVm

E= kqr2 wherer= r1→-r0→=4i^+5j^-i^-j^=3i^+4j^= 5 m⇒E=9 x 109 x 25 x 10-65 x 5 =9 x 103 Vm-1=9kVm

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A point charge 25 μC is located in the XY-plane at the point of position vector $\vec{r_{0}} = (\hat{i} + \hat{j}) m$ . What is the magnitude of electric field at the point of position vector $\vec{r_{1}} = (4 \hat{i} + 5 \hat{j}) m$ ?

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Can't be determined

900 V/m

9 kV/m

90 V/m

answer is B.

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$E = \frac{kq}{r^{2}}$ where

$r = |\vec{r_{1}} - \vec{r_{0}}| = |4 \hat{i} + 5 \hat{j} - \hat{i} - \hat{j}| = |3 \hat{i} + 4 \hat{j}| = 5 m$

$\Rightarrow E = \frac{9 x 10^{9} x 25 x 10^{- 6}}{5 x 5} = 9 x 10^{3} {Vm}^{- 1} = 9 \frac{kV}{m}$

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