Question

a electric flux of 6.5*10^3 is found to be linked with a sphere due to some charge inside it. calculate the magnitude of charge

Answer 1

Given:

Electric flux passing through sphere, Φ=6.5×10³Nm²/C

To Find:

Magnitude of charge

Solution:

We know that,

• According to Gauss's Law,

$\pink{\underline{\boxed{\bf{\phi=\dfrac{Q}{\varepsilon_{\circ}}}}}}$

where,

Φ is net electric flux through Gaussian surface

Q is magnitude of charge enclosed by Gaussian surface

ε₀ is absolute permittivity whose value is 8.85 x 10⁻¹² F/m

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Firstly, let us consider the sphere as Gaussian surface

Let the magnitude of charge inside sphere be Q

So,

$\longrightarrow\rm{\phi=\dfrac{Q}{\varepsilon_{\circ}}}$

$\longrightarrow\rm{6.5\times10^{3}=\dfrac{Q}{8.85\times10^{-12}}}$

$\longrightarrow\rm{6.5\times10^{3}\times8.85\times10^{-12}=Q}$

$\longrightarrow\rm{Q=6.5\times10^{3}\times8.85\times10^{-12}}$

$\longrightarrow\rm\green{Q=57.525\times10^{-9}\ C}$

Hence, the magnitude of charge  inside sphere is 57.525×10⁻⁹C.