Solution:

Ans. (i) Derivation of expression for electric field.

(ii) Finding point on line joining charges 2

[CBSE Marking Scheme, 2017]

Detailed Answer :

(i) Try yourself, Similar to Q. 1 (i), Short Answer Type Questions-II.

(ii) Two point charges $q_{A}=+4 \mu \mathrm{C}$ and $q_{B}=+1 \mu \mathrm{C}$ are located $2 \mathrm{~m}$ apart in air as shown. $P$ be a point, at a distance $x$ from $A$, where electric field of the system is zero.

Distance between the two charges, $A B=2 \mathrm{~m}$

$\therefore \quad \mathrm{PA}=x \mathrm{~m}, \mathrm{~PB}=(2-x) \mathrm{m}$

Electric field at point $P$ caused by $+4 \mu \mathrm{C}$ charge,

$$ E_{1}=\frac{4 \times 10^{-6}}{4 \pi \varepsilon_{0}(P A)^{2}} \mathrm{~N} / \mathrm{C} \text { along } P A \quad 1 / 2 $$

Magnitude of electric field at point $P$ caused by $+1 \mu \mathrm{C}$ charge

$E_{2}=\frac{+1 \times 10^{-6}}{4 \pi \varepsilon_{0}(P B)^{2}} \mathrm{~N} / \mathrm{C}$ along $P B$

Now, $\frac{4 \times 10^{-6}}{4 \pi \varepsilon_{0}(x)^{2}}=\frac{+1 \times 10^{-6}}{4 \pi \varepsilon_{0}(2-x)^{2}}$

(Since net electric field at $P$ is zero) $\mathbf{1}$ $(2-x)^{2} 4=(x)^{2} 1$

$16-16 x+3 x^{2}=0$

$4(4-x)-3 x(4-x)=0$

$(4-3 x)(4-x)=0$

$x=\frac{4}{3}, x=4$, possible point on the line will be $x=\frac{4}{3} 1 / 2$

AI Q. 3. (a) Derive an expression for the electric field $E$ due to a dipole of length ’ $2 a^{\prime}$ at a point distant $r$ from the centre of the dipole on the axial line.

(b) Draw a graph of $E$ versus $r$ for $r»a$.

(c) If this dipole were kept in a uniform external electric field $E_{0}$, diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.

R [O.D. Set III 2017]

[Topper’s Answer 2017]