Question

In the diagram, q1 = +4.88*10^-8 C.

What is the potential difference when
you go from point A to point B?
Include the correct sign, + or - .
B
0.538 m
1.36 m
91 |
(Hint: Does V go up or down when you go
from B to A?) (Unit = V)

Answer 1

The answer is 492.87.

Correct on Acellus

Answer 2

ΔV = 1139.3 V = 1.139 KV (+ve sign shows V goes up)

Step-by-step explanation:

The potential difference while moving from point A to Point B is given as follows:

`\Delta V = V_B-V_A`

where,

ΔV = potential difference from A to B = ?

`V_A` = Potential at point A =
`(kq)/(r_A)`

`V_B` = Potential at point B =
`(kq)/(r_B)`

Therefore,

`\Delta V = (kq)/(r_B)-(kq)/(r_A)\\\\\Delta V = kq((1)/(r_B)-(1)/(r_A))`

where,

k = Colomb's Constant = 9 x 10⁹ N.m²/C²

q = magnitude of charge = 4.88 x 10⁻⁸ C

`r_A` = distance of point A from charge = 1.36 m

`r_B` = distance of point B from charge = 0.538 m

Therefore,

`\Delta V = (9\ x\ 10^9\ N.m^2/C^2)(4.88\ x\ 10^(-8)\ C)((1)/(0.538\ m)-(1)/(1.36\ m))\\\\\Delta V = (439.2 N.m^2/C)(2.59\ /m)`

ΔV = 1139.3 V = 1.139 KV (+ve sign shows V goes up)