1. The electric field is uniform in a region of space, being equal to 500 N/C an

Question

1. The electric field is uniform in a region of space, being equal to 500 N/C and pointing to the east. Consider three points A, B and C in this region. The point A is to the west of B, and the distance between A and B is 8.0cm. The point C is north of B, and the distance between B and C is 4.0cm. Find

(a) the work done by the electric field when a point charge 3.0C moves from A to B,

(b) the potential difference VA -VB

(c) the work done by the electric field when the same charge moves from A to C, (c)

the potential difference VA -VC .

(d) Find the work done against the electric field in moving an electron from A to C

Explanation / Answer

Ekectric field is in the direction of east.

a) A and B are separated in the electric field direction.. Both points has different potential. That means there will be work done by electric field = q (VB-VA) = 3* [-500*0.08] = 120 Joule

b) Va-Vb = 500*0.08 = 40 Volt [ Potential decreases in the direction of electric field]

c) Electric field is conservative field hence it will give me same result irrelvant of the path i choosed.
Hence to move from A to c , i am first going A to B and then B to C.

Perpendicular to electric field plane it equipotential hence work done will be zero from B to C.
Hence Work done from A to C = work done from A to B = q (VB-VA) = 3* [-500*0.08] = 120 Joule

B and C are equipotential hence - Va- Vc = Va-Vb = 500*0.08 = 40 Volt [ Potential decreases in the direction of electric field

D )= work done against the electric field = - [ (-e) [ Va-Vc)] = 40 ev

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