Two point charges q_1 and q_2 (=q_1/2) are pl | vedclass

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(A) The electric field due to charge $q_1$ at point $A(0, 1)$ at point $P(1, 1)$ is directed along the positive $x$-axis.The distance $AP = 1$. Thus,$

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$	an^{-1}(1/2)$

Vedclass · Answer

(A) The electric field due to charge $q_1$ at point $A(0, 1)$ at point $P(1, 1)$ is directed along the positive $x$-axis.The distance $AP = 1$. Thus,$E_A = \frac{k q_1}{1^2} = k q_1$,where $k = \frac{1}{4 \pi \varepsilon_0}$.The electric field due to charge $q_2 = q_1/2$ at point $B(1, 0)$ at point $P(1, 1)$ is directed along the positive $y$-axis.The distance $BP = 1$. Thus,$E_B = \frac{k (q_1/2)}{1^2} = \frac{k q_1}{2}$.The resultant electric field vector at $P$ has components $E_x = E_A = k q_1$ and $E_y = E_B = \frac{k q_1}{2}$.The angle $	heta$ that the resultant electric field makes with the $x$-axis is given by $	an 	heta = \frac{E_y}{E_x} = \frac{E_B}{E_A} = \frac{k q_1 / 2}{k q_1} = \frac{1}{2}$.Therefore,$	heta = 	an^{-1}(1/2)$.

Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta$ with the $x$-axis,then the angle $\theta$ is

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