4. An electric field of magnitude 5.25 N/C points due east at a certain location

Question

4. An electric field of magnitude 5.25 N/C points due east at a certain location. Find the magnitude and direction of the force on a -6.55 µC charge at this location.

5. The potential difference between the accelerating plates of a TV set is about 20 kV. If the distance between the plates is 1.7 cm, find the magnitude of the uniform electric field in the region between the plates.

6. Find the electric field 0.2 m from a particle with a charge of +2 x 10-4 C. (ke = 8.99 x 109
Nm2/C2.)

Explanation / Answer

1)

|F| = |Eq| = (5.25 N/C) (6.55 x 10^(-6) C) = 3.43875x10^-5 newtons,
Direction = west (opposite of E field, because the charge is negative)

2)

Use E = V/d
E = 20 x10^3/1.7x 10^-2
E = 1.17 x 10^6 N/C

3)

Electric field E = kQ/r2

E = ((8.99x109)x(2x10-4))/(0.22)

E = 4.49x106 N/C

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