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The electric field at (30,30) cm due to a charge of -8nC at the origin in N/C

Saher katchi , 6 Years ago

Grade 12

1 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

Distance between origin and point (30,30)cm or (0.3,0.3)m is $d=sqrt{0.3^2+0.3^2}=0.3\sqrt{2}m$

Therefore, electric field at the point= $\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{d^2}=9\times 10^9\times \dfrac{8\times 10^{-9}}{0.18}N=400N$ towards the origin.

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