2018 A/L Physics Past Paper · Question 23 (English Medium)

Answer: (5) $\psi_R > +\psi$

Given information

We know: a net positive charge $Q$ is inside the closed surface $S$.

The flux through the part $A$ is $-\psi$. The minus sign means the flux comes in through $A$.

To see it with numbers, take

$$\psi = 1,\; \frac{Q}{\varepsilon_0} = 2$$

What the question asks

$A$ and the rest together make the whole surface $S$.

So their two fluxes add up to the total flux of $S$.

We want to compare $\psi_R$ with $\psi$.

Relevant theory

The total normal electric flux crossing a closed surface of any shape surrounding a charge Q situated in a medium of permittivity

For the sake of convenience, it is considered that the electric field intensity is the number of electric flux lines crossing normally a unit area of the field.

Required equations

$\Phi_{S} = \frac{Q}{\varepsilon_0}$

$\Phi_S = (-\psi) + \psi_R$

Applying it to this question

Flux going out of the surface is positive.

Flux coming in is negative.

So the $-\psi$ through $A$ is flux coming in.

The charge inside is positive. So the total flux out of $S$ is positive.

$$\Phi_S = \frac{Q}{\varepsilon_0} > 0$$

Set this equal to the flux of $A$ plus the flux of the rest.

Reasoning and calculation

$-\psi + \psi_R = \frac{Q}{\varepsilon_0}$

$\psi_R = \psi + \frac{Q}{\varepsilon_0}$

$\psi_R > +\psi$

Checks

$psi + q_eps$

Check with field lines.

The lines that come in through $A$ must go out somewhere else.

The lines from the charge inside also go out.

Both go out through the rest. So the flux there is more than $\psi$.

Conclusion

$\psi_R = \psi + \frac{Q}{\varepsilon_0} > \psi$

$$\psi_R = \psi + \frac{Q}{\varepsilon_0}$$

Here $Q > 0$. So $\psi_R > +\psi$.

The answer is option (5).