Question

In an electrostatic field, path 1 between points A and B is twice as long as path 2. The electrostatic work done on a negatively charged particle that moves from A to B along path 1 is W1.

a) How much work is done on this particle if it later goes from A to B along path 2? (Express your answer in terms of W1.)

Answer 1

Answer:

Electrostatic field is a conservative field. That is work done when a charge is moved between two points is independent of path taken. Hence answer is W1

Answer 2

The work done on the particle as it later goes from A to B along path 2 is 2W₁.

The given parameters;

• part 1: distance between A and B = 2r
• part 2: distance between A and B = r

The work done in moving a charge from one point to another is calculated by applying Coulomb's law;

$V = \frac{kq}{r}$

The work done in moving the charge between the two points along path 1 is calculated as;

$W_1 = \frac{kq}{2r}$

The work done in moving the charge between the two points along path 2 is calculated as;

$W_2 = \frac{kq}{r} \\\\divide \ both \ sides \ by\\\\\frac{W_2}{2} = \frac{kq}{2r}\\\\\frac{W_2}{2} = W_1\\\\W_2 = 2W_1$

Thus, the work done on the particle as it later goes from A to B along path 2 is 2W₁.

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