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(NONE) Let the distance of each charge from the origin be $r$. The electric field at the origin is $\vec{E} = \sum \frac{kQ_i}{r^2} \hat{r}_i$. The fo

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$F_1  >  F_2  >  F_3  >  F_4$

Vedclass · Answer

(NONE) Let the distance of each charge from the origin be $r$. The electric field at the origin is $\vec{E} = \sum \frac{kQ_i}{r^2} \hat{r}_i$. The force is $\vec{F} = q_0 \vec{E}$.Figure-$1$: Charges are $2Q$ at $A(-r,0)$,$-3Q$ at $B(r,0)$,and $5Q$ at $C(0,r)$.$\vec{E}_1 = \frac{k}{r^2} [(-2Q)\hat{i} - (-3Q)\hat{i} - 5Q\hat{j}] = \frac{k}{r^2} [Q\hat{i} - 5Q\hat{j}]$. Magnitude $E_1 = \frac{kQ}{r^2} \sqrt{1^2 + (-5)^2} = \frac{kQ}{r^2} \sqrt{26} \approx 5.1 \frac{kQ}{r^2}$.Figure-$2$: Charges are $2Q$ at $A(-r,0)$,$3Q$ at $B(r,0)$,and $5Q$ at $C(0,r)$.$\vec{E}_2 = \frac{k}{r^2} [(-2Q)\hat{i} - (3Q)\hat{i} - 5Q\hat{j}] = \frac{k}{r^2} [-5Q\hat{i} - 5Q\hat{j}]$. Magnitude $E_2 = \frac{kQ}{r^2} \sqrt{(-5)^2 + (-5)^2} = \frac{kQ}{r^2} \sqrt{50} \approx 7.07 \frac{kQ}{r^2}$.Figure-$3$: Charges are $2Q$ at $A(-r,0)$,$-2Q$ at $B(r,0)$,$5Q$ at $C(0,r)$,and $5Q$ at $D(0,-r)$.$\vec{E}_3 = \frac{k}{r^2} [(-2Q)\hat{i} - (-2Q)\hat{i} - 5Q\hat{j} + 5Q\hat{j}] = 0$.Figure-$4$: Charges are $2Q$ at $A(-r,0)$,$2Q$ at $B(r,0)$,$5Q$ at $C(0,r)$,and $5Q$ at $D(0,-r)$.$\vec{E}_4 = \frac{k}{r^2} [(-2Q)\hat{i} - (2Q)\hat{i} - 5Q\hat{j} + 5Q\hat{j}] = \frac{k}{r^2} [-4Q\hat{i}]$. Magnitude $E_4 = 4 \frac{kQ}{r^2}$.Comparing magnitudes: $E_2 (7.07)  >  E_1 (5.1)  >  E_4 (4)  >  E_3 (0)$.Thus,$F_2  >  F_1  >  F_4  >  F_3$.

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