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Physics · 2023 · 5 marks
CBSE 2023 · Region 1 · Set 1 · Q31
(a) (i) State Coulomb's law in electrostatics and write it in vector form, for two charges. (ii) 'Gauss's law is based on the inverse-square…
(i)
State Coulomb's law in electrostatics and write it in vector form, for two charges.
(ii)
'Gauss's law is based on the inverse-square dependence on distance contained in the Coulomb's law.' Explain.
(iii)
Two charges A (charge q ) and B (charge $\displaystyle 2$ q ) are located at points $\displaystyle (0,0)$ and $\displaystyle (a, a)$ respectively. Let $\displaystyle \hat{\mathrm{i}}$ and $\displaystyle \hat{\mathrm{j}}$ be the unit vectors along x -axis and y -axis respectively. Find the force exerted by A on B , in terms of $\displaystyle \hat{\mathrm{i}}$ and $\displaystyle \hat{\mathrm{j}}$.
(i)
Derive an expression for the electric field at a point on the equatorial plane of an electric dipole consisting of charges q and -q separated by a distance 2a.
(ii)
The distance of a far off point on the equatorial plane of an electric dipole is halved. How will the electric field be affected for the dipole?
(iii)
Two identical electric dipoles are placed along the diagonals of a square ABCD of side $\displaystyle \sqrt{2} \mathrm{~m}$ as shown in the figure. Obtain the magnitude and direction of the net electric field at the centre (O) of the square.
Marking-scheme solution
i) Force between two point charges varies inversely with the square of distance between the charges and is directly proportional to the product of magnitude of the two charges and acts along the line joining the two charges.
\[\overrightarrow{F_{12}}=\frac{1}{4 \pi \epsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}{ }^{2}} \widehat{\mathrm{r}_{12}}
\]
Alternatively
\(\displaystyle \overrightarrow{F_{12}}=\frac{1}{4 \pi \epsilon_{o}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}_{12}{ }^{3}} \overrightarrow{\mathrm{r}_{12}}\)
Where \(\displaystyle \overrightarrow{\mathrm{r}_{12}}\) is a vector from charge \(\displaystyle \mathrm{q}_{2}\) to charge \(\displaystyle \mathrm{q}_{1}\).ii) In derivation of Gauss's law, flux is calculated using Coulomb's law and surface area. Here coulomb's law involves \(\displaystyle \frac{1}{\mathrm{r}^{2}}\) factor and surface area involves \(\displaystyle \mathrm{r}^{2}\) factor. When product is taken, the two factors cancel out and flux becomes independent of r.
iii) 
\(\displaystyle \vec{\mathrm{r}}=\overrightarrow{A B}=a \hat{\mathrm{i}}+a \hat{\mathrm{j}}\)
\(\displaystyle \mathrm{r}=|\overrightarrow{A B}|=\sqrt{a^{2}+a^{2}}=\sqrt{2} a\)
\(\displaystyle \vec{F}=\frac{1}{4 \pi \epsilon_{o}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)
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