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

The situation is represented in the given figure. O is the mid-point of line AB.

• A+3µC
• B -3µC

Distance between the two charges, AB 20cm, AO = OB = 10cm

Net electric field at point O = E Electric field at point O caused by +3 C charge,

E₁ 3 × 10 6 4π €₁ (AO)² 3 x 10-6 4TT €o (10 × 10-2)2N/C along OB

Where,

Eo= Permittivity of free space 1 4π Eo 9 × 10ºNm²C-2²

Magnitude of electric field at point O caused

by - 3 C charge,

E₂ -3 x 10-6 =

4π €₁ (OB)² 3 x 10-6 4π Eo (10 x 10-2)² N/C

along OB, E = E₁ + E₂2

= 2 × (9 × 10⁹) × (10 × 10-2)²

[the value is multiplied with 2]

= 5.4 x 106 N/C along OB

Therefore, the electric field at mid-point O is 5.4 x 106 N/C along OB.

(b) A test charge of amount q = 1.5 × 10-⁹℃ is placed at mid-point O.q=1.5 × 10 °C

Force experienced by the test charge

= FF =qE

= 1.5 x 10-9

= 8.1 x 10-³N

The force is along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Hence, the force experienced by the test charge is 8.1 x 10-³ Nalong OA.

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