Question

In the diagram, q1 = +3.25*10^-9 C.

What is the potential difference when
you go from point B to point A?
q1
A
B
0.150 m
0.250 m

Answer 1

121.7

Step-by-step explanation:

credit to the comment above^

Answer 2

ΔV = 78 V

Step-by-step explanation:

The potential at a point due to a point charge is given as:

`V = (kq)/(r)`

where,

V = Potential at the point

k = Colomb's Constant = 9 x 10⁹ Nm²/C²

q = charge = 3.25 x 10⁻⁹ C

r = distance

AT POINT A:

`V_A = ((9\ x\ 10^9\ Nm^2/C^2)(3.25\ x\ 10^(-9)\ C))/(0.15\ m) \\\\V_A = 195\ V`

AT POINT B:

`V_B = ((9\ x\ 10^9\ Nm^2/C^2)(3.25\ x\ 10^(-9)\ C))/(0.25\ m) \\\\V_B = 117\ V`

Now, the potential difference will be:

`\Delta V = V_A-V_B = 195\ V - 117\ V`

ΔV = 78 V