Question

Four charges QA = 2μC, QB = -13μC, QC = 3μC and QD = -4μC are placed in a hollow sphere. Calculate electric flux through it.

Answer 1

Answer:

             $\boxed{\sf \phi=\dfrac{-12\times 10^{-6}}{\in_{o}}}$

Explanation:

$\rule{300}{1.5}$

From the formula we know that,

$\large\bigstar\;\underline{\boxed{\sf \phi=\displaystyle\oint\sf E.dA=\dfrac{q}{\in_{o}}}}$

Here,

• Φ Denotes Electric flux.
• q Denotes the charge enclosed.
• ε₀ Denotes permittivity of free space.

Now, we are given that four charges are placed in hollow sphere.

Therefore, simplifying it,

$\longmapsto\sf \phi=\dfrac{q}{\in_{o}}\\\\\\\\\longmapsto\sf \phi=\dfrac{Q_{A}+Q_{B}+Q_{C}+Q_{D}}{\in_{o}}\\\\\\\\\longmapsto\sf \phi=\dfrac{2\mu C+\bigg(-13\mu C\bigg)+3\mu C+\bigg(-4\mu C\bigg)}{\in_{o}}\\\\\\\\\longmapsto\sf \phi=\dfrac{\bigg(2-13+3-4\bigg)\times 10^{-6}}{\in_{o}}\\\\\\\\\longmapsto\sf \phi=\dfrac{\bigg(5-17\bigg)\times10^{-6}}{\in_{o}}\\\\\\\\\longmapsto\sf \phi=\dfrac{-12\times10^{-6}}{\in_{o}}\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf \phi=\dfrac{-12\times10^{-6}}{\in_o}}}}}$

Hence, we got the electric flux through the sphere.

Note:

• You can simplify more by substituting the value of ε₀ (8.85 × 10⁻¹²) which will give a value of - 1.35 × 10⁶ Nm²/C.

$\rule{300}{1.5}$

Answer 2

Answer:

-12 × 10^-3

Explanation:

is the correct answer