CBSE Class 12 Physics 2024: question 17(a)

Hints

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Step-by-step solution

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1. Each corner is at the same distance from the centre O: $r=\frac{30\sqrt{2}}{2}\text{ cm}=15\sqrt{2}\text{ cm}$, so $r^2=0.045{\text{m}}^2$.

   Diagram (in words): draw square ABCD with O at the centre.

   • ${\vec{F}}_{OA}$ and ${\vec{F}}_{OC}$ are repulsions, since $+1\mu$C acts on $+4\mu$C. They point along AO and CO, which are opposite directions.
   • ${\vec{F}}_{OB}$ and ${\vec{F}}_{OD}$ are attractions, since $-2\mu$C acts on $+4\mu$C. They point from O towards B and from O towards D, which are also opposite directions.
   1 mark

2. $F_{OA}=F_{OC}=\frac{9\times 10^9\times 4\times 10^{-6}\times 1\times 10^{-6}}{0.045}=0.8\text{N}$.

   These two forces are equal and opposite, so ${\vec{F}}_{OA}+{\vec{F}}_{OC}=0$.

   ½ mark

3. $F_{OB}=F_{OD}=\frac{9\times 10^9\times 4\times 10^{-6}\times 2\times 10^{-6}}{0.045}=1.6\text{N}$.

   These two forces are also equal and opposite, so ${\vec{F}}_{OB}+{\vec{F}}_{OD}=0$.

   Hence the net force on the $4\mu$C charge is $\vec{F}=0$.

   ½ mark

Answer: Net force on the 4 μC charge = 0

Revise the chapter

• Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)