Question

Three charges are enclosed inside a spherical closed surface. The net flux through the surface is −216 N · m2/C. If two of the charges inside the closed surface are 1.74 nC and 1.16 nC, determine the value (magnitude and sign) of the third charge.

Answer 1

q₃ = -4.81 nC

Step-by-step explanation:

We can use the Gauss Law here:

∅ = q/∈₀

where,

∅ = Net Flux = - 216 N.m²/C

q = total charge enclosed inside sphere = ?

∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²

Therefore,

- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²

q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)

q = - 1.91 nC

So, the total charge will be sum of all three charges:

q = q₁ + q₂ + q₃

- 1.91 nC = 1.74 nC + 1.16 nC + q₃

q₃ = - 1.91 nC - 1.74 nC - 1.16 nC

q₃ = -4.81 nC