Question

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

Answer 1

Answer:

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

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 = $\frac{kq}{r_A}$

$V_B$ = Potential at point B = $\frac{kq}{r_B}$

Therefore,

$\Delta V = \frac{kq}{r_B}-\frac{kq}{r_A}\\\\\Delta V = kq(\frac{1}{r_B}-\frac{1}{r_A})$

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,

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

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