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Question 1 of 10
1. Question
CorrectIncorrectHint
$\Phi_E = EA$
\[
[E] = \frac{[F]}{[q]}
\]\[
[F] = MLT^{-2}, \qquad [q] = AT
\]\[
[E] = \frac{MLT^{-2}}{AT}
= MLT^{-3}A^{-1}
\]Since
\[
[A] = L^2,
\]we get
\[
[\Phi_E] = [E][A]
\]\[
= \left(MLT^{-3}A^{-1}\right)(L^2)
\]\[
\boxed{[\Phi_E] = ML^3T^{-3}A^{-1}}
\] -
Question 2 of 10
2. Question
CorrectIncorrectHint
\[
\frac{1}{4\pi\epsilon_0}
= 9\times10^9\ \mathrm{N\,m^2\,C^{-2}}
\]\[
\boxed{\frac{1}{4\pi\epsilon_0}=9\times10^9\ \mathrm{N\,m^2\,C^{-2}}}
\] -
Question 3 of 10
3. Question
CorrectIncorrectHint
\[
\text{By Coulomb’s law,}
\]\[
F \propto \frac{1}{r^2}
\]Let the new separation be \(r’\).
\[
\frac{F’}{F}=\frac{r^2}{r’^2}
\]Given,
\[
F=25\,\mathrm{N}, \qquad F’=5\,\mathrm{N}
\]Therefore,
\[
\frac{5}{25}=\frac{r^2}{r’^2}
\]\[
\frac{1}{5}=\frac{r^2}{r’^2}
\]\[
r’^2=5r^2
\]\[
\boxed{r’=\sqrt{5}\,r}
\]Hence, the separation should be **increased to \(\sqrt{5}\) times the original separation**.
-
Question 4 of 10
4. Question
CorrectIncorrectHint
\[
F_e=\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}
\]\[
F_g=G\frac{m_em_p}{r^2}
\]Therefore,
\[
\frac{F_e}{F_g}
=
\frac{e^2}{4\pi\epsilon_0 Gm_em_p}
\]Using
\[
e=1.6\times10^{-19}\,\mathrm{C},
\quad
\frac{1}{4\pi\epsilon_0}=9\times10^9,
\]\[
G=6.67\times10^{-11},
\quad
m_e=9.1\times10^{-31}\,\mathrm{kg},
\quad
m_p=1.67\times10^{-27}\,\mathrm{kg},
\]we get
\[
\frac{F_e}{F_g}\approx1.2\times10^{39}.
\]Hence, the ratio is of the order of
\[
\boxed{10^{39}}.
\] -
Question 5 of 10
5. Question
CorrectIncorrectHint
\[
\boxed{\text{remains }e}
\]\[
\text{The electric charge of a proton is invariant and does not depend on its speed.}
\]\[
q=e=1.6\times10^{-19}\,\mathrm{C}
\] -
Question 6 of 10
6. Question
CorrectIncorrectHint
\[
\boxed{\text{Coulomb’s law}}
\]\[
F=\frac{1}{4\pi\epsilon_0}\frac{Q_1Q_2}{r^2}
\]\[
\therefore\quad F\propto\frac{Q_1Q_2}{r^2}
\] -
Question 7 of 10
7. Question
CorrectIncorrectHint
\[
F=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}
\]In a medium of dielectric constant \(K\),
\[
F’=\frac{1}{4\pi\epsilon_0K}\frac{q_1q_2}{r’^2}
\]Given,
\[
r’=\frac{r}{2}, \qquad K=2
\]Therefore,
\[
F’=\frac{1}{4\pi\epsilon_0(2)}
\frac{q_1q_2}{(r/2)^2}
\]\[
F’=\frac{1}{2}\times4
\left(\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\right)
\]\[
F’=2F
\]\[
\boxed{F’=2F}
\] -
Question 8 of 10
8. Question
CorrectIncorrectHint
\[
\boxed{\epsilon_0 = 8.85\times10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}}
\]\[
\text{or equivalently,}
\]\[
\boxed{\epsilon_0 = 8.85\times10^{-12}\ \mathrm{F\,m^{-1}}}
\] -
Question 9 of 10
9. Question
CorrectIncorrectHint
\[
\rho_v=\frac{Q}{V}
\]Since charge has SI unit \(\mathrm{C}\) and volume has SI unit \(\mathrm{m^3}\),
\[
[\rho_v]=\frac{\mathrm{C}}{\mathrm{m^3}}
\]\[
\boxed{\mathrm{C\,m^{-3}}}
\] -
Question 10 of 10
10. Question
CorrectIncorrectHint
\[
q = ne
\]where \(n\) is an integer and \(e\) is the elementary charge.
\[
\boxed{\text{Quantisation of electric charge}}
\]