Answer:
1. False
2. True
3. True
Explanation:
1- False —> The relation between electric potential and electric field is given such that

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.
2- True —> The electric field lines always cross the equipotential lines perpendicularly.
3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.
The first part of the question is:
Two point charges are placed on the x axis. (Attached file)The first charge, q1 = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.
Answer:
q3 = +0.3nc
Explanation:
Due to the vector symbols in the solution, I've decided to attach the explanation to this answer.
If you put a penny in each light spot the penny that the light is shining on will recive the most energy.