CBSE 2024 · Region 5 · Set 1 · Q17 · 2 marks
(a)Four point charges of $\displaystyle 1 \mu \mathrm{C},-2 \mu \mathrm{C}, 1 \mu \mathrm{C}$ and $\displaystyle -2 \mu \mathrm{C}$ are placed at the corners $\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C}$ and D respectively, of a square of side $\displaystyle 30$ cm . Find the net force acting on a charge of $\displaystyle 4 \mu \mathrm{C}$ placed at the centre of the square.OR 17. (b) Three point charges, $\displaystyle 1$ pC each, are kept at the vertices of an equilateral triangle of side $\displaystyle 10$ cm . Find the net electric field at the centroid of triangle.
(a)
Four point charges of $\displaystyle 1 \mu \mathrm{C},-2 \mu \mathrm{C}, 1 \mu \mathrm{C}$ and $\displaystyle -2 \mu \mathrm{C}$ are placed at the corners $\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C}$ and D respectively, of a square of side $\displaystyle 30$ cm . Find the net force acting on a charge of $\displaystyle 4 \mu \mathrm{C}$ placed at the centre of the square.
OR 17. (b) Three point charges, $\displaystyle 1$ pC each, are kept at the vertices of an equilateral triangle of side $\displaystyle 10$ cm . Find the net electric field at the centroid of triangle.
Marking-scheme solution
(a)
$\displaystyle O A=O B=O C=O D=r$
Net force on charge $\displaystyle 4 \mu \mathrm{C}$:
$\displaystyle \vec{F}=\vec{F}_{O A}+\vec{F}_{O B}+\vec{F}_{O C}+\vec{F}_{O D}$
$\displaystyle \vec{F}_{O A}=-\vec{F}_{O C} \Rightarrow \vec{F}_{O A}+\vec{F}_{O C}=0$
$\displaystyle \vec{F}_{O B}=-\vec{F}_{O D} \Rightarrow \vec{F}_{O B}+\vec{F}_{O D}=0$
$\displaystyle F=0$
$\displaystyle F_{O A}=F_{O C}=\frac{9 \times 10^{9} \times 4 \times 10^{-6} \times 1 \times 10^{-6}}{\left(15 \sqrt{2} \times 10^{-2}\right)^{2}}=0.8 \mathrm{~N}$
$\displaystyle F_{O B}=F_{O D}=1.6 \mathrm{~N}$
$\displaystyle F_{1}=F_{O A}-F_{O C}=0$
$\displaystyle F_{2}=F_{O B}-F_{O D}=0$
Net force $\displaystyle F=0$
$\displaystyle q_{A}=q_{B}=q_{C}$
$\displaystyle A O=B O=C O=r$
$\displaystyle E_{O A}=E_{O B}=E_{O C}$
$\displaystyle \vec{E}_{B C}=\vec{E}_{O B}+\vec{E}_{O C}$
$\displaystyle E_{B C}=\sqrt{E_{O B}^{2}+E_{O C}^{2}+2 E_{O B} E_{O C} \cos 120^{\circ}}$
$\displaystyle E_{B C}=E_{O B}$, $\displaystyle \vec{E}_{O A}=-\vec{E}_{B C}$
Net electric field $\displaystyle \vec{E}_{O}=\vec{E}_{O A}+\vec{E}_{B C}$
$\displaystyle E_{O}=0$
$\displaystyle E_{O A}=E_{O B}=E_{O C}$
Electric field vectors make an angle of $\displaystyle 120^{\circ}$ with each other. They make a closed polygon, so the vector sum of all electric field vectors will be zero.
$\displaystyle E=0$
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.