When a point charge with q = - 2 × 10-6c is removed from a point in the electric field and moves from point a to point B, the electrostatic force is overcome and the work is done by 6 × 10-6c When a point charge with q = - 2 × 10-6c is removed, when it moves from point a to point B in the electric field, it overcomes the electrostatic force to do 6 × 10-4 J, and when it moves from point B to point C, it does 9 × 10-4 J (1) With B as the zero potential energy point, what is the electric potential energy EPA of the charge at point a? (2) With C as the zero potential energy point, what is the electric potential energy EPA of charge at point a?

When a point charge with q = - 2 × 10-6c is removed from a point in the electric field and moves from point a to point B, the electrostatic force is overcome and the work is done by 6 × 10-6c When a point charge with q = - 2 × 10-6c is removed, when it moves from point a to point B in the electric field, it overcomes the electrostatic force to do 6 × 10-4 J, and when it moves from point B to point C, it does 9 × 10-4 J (1) With B as the zero potential energy point, what is the electric potential energy EPA of the charge at point a? (2) With C as the zero potential energy point, what is the electric potential energy EPA of charge at point a?

(1) Taking B as the zero potential energy point, from a to B, the electrostatic force does negative work and the potential energy increases, so the potential energy at point a is negative. EPA = 6 × 10-4 J
(2) Taking point C as the zero electric potential energy point, from B to C, the electrostatic force does positive work and the electric potential energy decreases, so EPB = 9 × 10-4 J
From a to B, the electrostatic force does negative work and the electric potential energy increases, so EPA = epb-6 × 10-4 J = 3 × 10-4 J

In the electric field, a point charge of - 6 × 10-8c is moved from point a to point B, and the work done by the electric field force is - 3 × 10-5j. When the charge is moved from point B to point C, the work done by the electric field force is 4.5 × 10-5j, and the potential difference between a and C is calculated

According to w = Qu: WAC = WAB + WBC = - 3 × 10-5 + 4.5 × 10-5j = 1.5 × 10-5j = Quac: UAC = wacq = 1.5 × 10 − 5 − 6 × 10 − 8V = - 250V A: the potential difference between AC is - 250V

The point charge of - 7 power C with charge amount of 2 * 10 is moved from point a with potential of 400V to point B with potential of - 100V, and the work done by electric field force () J,
How electric potential energy changes: () (increase / decrease / constant)

WAB = quab = 1.0x10 ^ - 4J potential energy decrease