Question 194: Electrostatics & Capacitance

Question

If $\epsilon_0$ denotes the permittivity of free space and $\Phi_E$ is the flux of the electric field through the area bounded by the closed surface, then dimensions of $\left(\epsilon_0 \frac{d \phi_E}{d t}\right)$ are that of :

Accepted Answer

We can use GaussÃ¢ÂÂs law to analyze the dimensions. GaussÃ¢ÂÂs law states that
$$ \Phi_E = \frac{q}{\epsilon_0}, $$
where
$$\Phi_E$$ is the electric flux,
$$q$$ is the electric charge,
$$\epsilon_0$$ is the permittivity of free space.
To find the dimensions of $$\epsilon_0 \frac{d\Phi_E}{dt}$$, follow these steps:
Differentiate GaussÃ¢ÂÂs law with respect to time:
$$    \frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0}\frac{dq}{dt}.    $$
Multiply the equation by $$\epsilon_0$$:
$$    \epsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}.    $$
Recognize that the time rate of change of charge, $$\frac{dq}{dt}$$, is by definition the electric current.
Thus, the dimensions of $$\epsilon_0 \frac{d\Phi_E}{dt}$$ are those of electric current.
The correct answer is: Option C, electric current.

Answer

electric charge

Answer

electric field

Answer

electric current

Answer

electric potential

Electrostatics & Capacitance

If \epsilon_0 denotes the permittivity of free space and \Phi_E is the flux of the electric field through the area bounded by the closed surface, then dimensions of \left(\epsilon_0 \frac{d \phi_E}{d t}\right) are that of :

Choose the correct answer: