Question

cosider rectangle PQRS of size 20cm and 156cm with charges 0.000002C, 0.000004C and 0.000009C at PQ and S respectively. determine the resultant electric field intensity at fourth corner R of the rectangle due to the changes PQS

Answer 1

Answer on Question #70835 – Physics – Field Theory

Consider rectangle PQRS of size $20\mathrm{cm}$ and $156\mathrm{cm}$ with charges $0.000002C$, $0.000004C$ and $0.000009C$ at PQ and S respectively. Determine the resultant electric field intensity at fourth corner R of the rectangle due to the changes PQS

Solution. Let PS=20cm.

$$E_{QR} = \frac{1}{4\pi\varepsilon_0} \frac{q_Q}{R_{QR}^2} \approx 898,755 \frac{N}{C},$$ $$E_{SR} = \frac{1}{4\pi\varepsilon_0} \frac{q_S}{R_{SR}^2} \approx 33,238 \frac{N}{C},$$ $$\angle PRS = \theta = \arctan 20/156 \approx 7.3{}^\circ$$ $$E_{PR} = \frac{1}{4\pi\varepsilon_0} \frac{q_P}{R_{PR}^2} \approx 7,267 \frac{N}{C}.$$ $$E_{tot} = \sqrt{\left(E_{QR} + E_{PR} \sin \theta\right)^2 + \left(E_{SR} + E_{PR} \cos \theta\right)^2} = 900,587 \frac{N}{C} \approx 9 \times 10^5 \frac{N}{C}$$

Answer. $E_{tot} = 9 \times 10^5 \frac{N}{C}$.

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