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Sagot :
(a) The magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
(b) The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
(c) The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
(d) The potential difference between A and B is 56.7 V.
Electric force on the sphere
The electric force on the sphere is calculated as follows;
F = Eq
where;
- E is electric field
- q is the charge
F = 300 x (44 x 10⁻⁹)
F = 1.32 x 10⁻⁵ N
The direction of the force is towards the right.
Work done on the sphere
W = Fd
W = 1.32 x 10⁻⁵ N x 0.189 m
W = 2.5 x 10⁻⁶ J
Change of the electric potential energy
The change in the electric potential energy (in J) as the sphere moves from A to B is equal to work done in moving the charge = 2.5 x 10⁻⁶ J.
Potential difference between A and B
VB − VA = Ed
VB − VA = 300 N/C x 0.189 m
VB − VA = 56.7 V
Thus, the magnitude of the force is 1.32 x 10⁻⁵ N and direction of the electric force on the sphere towards the right.
The work done on the sphere by the electric force as it moves from A to B is 2.5 x 10⁻⁶ J.
The change of the electric potential energy as the sphere moves from A to B is 2.5 x 10⁻⁶ J.
The potential difference between A and B is 56.7 V.
Learn more about potential difference here: https://brainly.com/question/9060304
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