Explanation:
It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.
So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.
U = qV
Since, the electric potential is directly proportional to the charge. Hence, when 0.2 tends to replaced by 0.4 then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.
Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.
Matter. I don't really know how to explain it. Sorry. But anyways, Hope this helps!
Explanation:
If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :
F = q E
Hence, this is the required solution.
F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.
Or it could be 15 if both top and bottom are 25. 10+15= 25.