A uniform electric field has magnitude E and is directed in the negative x direction. The potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) is 240 V. Required:
a. Which point, a or b, is at the higher potential?
b. Calculate the value of £.
c. A negative point charge q = —0.200 μC is moved from 5 to a. Calculate the work done on the point charge by the electric field.

a)

b is at higher potential

b)

E = 800 V/m

c)

W = -48.0 μ J

Explanation:

Given:

electric field has magnitude E and is directed in the negative x direction

potential difference between point a (at x = 0.60 m) and point b (at x = 0.90 m) = 240 V

a)

Direction of the electric field is in the negative x direction so it is away from the positive charge and the more distance away from direction, the lesser the potential gets. So b is at the higher potential.

b)

When the magnitude of electric field is constant then the potential difference between two points in the field is given by:

Vab = Ed

where

Vab is the potential difference between two points a and b in the field

E is electric field magnitude

d is the distance between two points

Compute distance between two points a and b

d = 0.90 - 0.60

d = 0.30

d = 0.3 m

Since potential difference between two points in the field is given which is 240 V

Compute E:

The above formula becomes:

Vab = Ed

E = Vab/d

  = 240/0.3

E = 800 V/m

c)

The negative charge moved from the higher potential to low so work done on the point charge by electric field is negative.

W = Fd

Electric field = E = F/q

where

E is electric field magnitude

F is electric force on q

q is point charge magnitude

The charge q is either positive or negative and when charge is positive the directions of E and F are same otherwise opposite in case of negative charge.

So to calculate the work done on the point charge by the electric field:

E = F/q

F = Eq

so

W = Fd = Eqd

Now putting the values:

E = 800 V/m

q = - 0.200 μC

d = 0.3 m

W = Eqd

W = 800 * - 0.200 * 0.3

W = 800 * - 0.2 * 10⁻⁶ * 0.3

W = - 48 * 10⁻⁶

W = - 4.8 * 10⁻⁵ J

W = -48.0 μJ

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speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.