Answer:
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
a. The point which is at a higher potential is point a.
b. The magnitude (E) of the uniform electric field is equal to 800 V/m.
c. The work done on the point charge by the electric field is equal to -48 microjoules.
Given the following data:
a. Since the uniform electric field with a magnitude (E) is directed in the negative x direction, the point at a higher potential is point a because the electric field is not close to the positive charge and it is further away from point b by 0.90 meter. Also, the potential difference of an electric field is inversely proportional to the square of the distance from a point charge.
b. To calculate the magnitude (E) of the uniform electric field:
At a constant potential difference, the magnitude (E) of this electric field is given by the formula:
[tex]V_{ab} = Ed[/tex]
Where:
First of all, we would determine the distance (d).
[tex]d = d_b - d_a\\\\d = 0.90 -0.60[/tex]
Distance, d = 0.30 meter.
Substituting the given parameters into the formula, we have;
[tex]240 = 0.3E\\\\E=\frac{240}{0.3}[/tex]
E = 800 V/m
c. To calculate the work done on the point charge by the electric field:
Mathematically, the work done on a point charge in an electric field is given by the formula:
[tex]W = Fd = Eqd\\\\W = 800 \times (-0.200 \times 10^{-6}) \times 0.3\\\\W = -240 \times 0.200 \times 10^{-6}\\\\W = -48\times 10^{-6}\;Joules[/tex]
Note: 1 microjoules = [tex]1\times 10^{-6}[/tex] Joules
Work done = -48 microjoules.
Read more on an electric field here: https://brainly.com/question/23153766
