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Chapter 4: Q24E (page 715)

A particle has charge -5.00 nC. (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 m directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 N/C?

Short Answer

Expert verified
  1. Magnitude and direction of the electric field applied by the particle is E=720N/C, and the direction is downwards to the particle.
  2. Distance is 1.94m.

Step by step solution

01

Step 1:

As given the distance of the point is r=0.250, and the charge of the particle is negative, so the distance of the electric field is downwards.

For electric field:

E=14πϵ0⋅|q|r2

Here;

14πϵ0=9×109Nm2/C2,r=0.250m

Putting all these values,

E=14πϵ0⋅|q|r2=9×109×−5×10−9(0.25)=720N/C

Therefore, the Magnitude and direction of the electric field applied by the particle is E=720N/C, and the direction is downwards to the particle.

02

Step 2:

AsE∝1r2,

Therefore, the relation between two states of electric field

E1E2=r22r12r2=r1E1/E2=0.250720/12=1.94m

Hence, Distance is 1.94m.

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