Answer:
(a) increase
Explanation:
On a line graph; we have the x-axis to the positive side and the negative side .In a positive x-axis direction, the force is usually positive, vice versa the negative side as well.
The change in the potential energy of a charge field-system can be given as:
[tex]\delta U= -q(EdsCos \theta)[/tex]
where;
q = positive test charge
E = Electric field
ds = displacement between thee charge positions
θ = Angle between the electric field and the displacement.
Given that:
Charge of the particle = -q
displacement = (60.0 -20.0)cm = 40.0 cm
θ = 0
Replacing our values in the above equation, we have:
[tex]\delta U = -(-q)(60Cos 0)[/tex]
[tex]\delta U = qEds[/tex]
Since the potential energy of the system is positive, therefor the electric potential energy also increases.
