If the net electric field at point P along th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the distance from the charges to point $P$ be $r_1$ and $r_2$. From the geometry,$r_1 = \sqrt{4^2 + 2^2} = \sqrt{20} 	ext{ cm}$ and $r_2 = \s

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let the distance from the charges to point $P$ be $r_1$ and $r_2$. From the geometry,$r_1 = \sqrt{4^2 + 2^2} = \sqrt{20} 	ext{ cm}$ and $r_2 = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 	ext{ cm}$.The electric field due to $q_2$ at $P$ is $E_1 = \frac{K q_2}{r_1^2} = \frac{K q_2}{20}$. The component along the $Y$-axis is $E_{1y} = E_1 \cos \beta = \frac{K q_2}{20} \cdot \frac{4}{\sqrt{20}}$.The electric field due to $q_3$ at $P$ is $E_2 = \frac{K q_3}{r_2^2} = \frac{K q_3}{25}$. The component along the $Y$-axis is $E_{2y} = E_2 \cos 	heta = \frac{K q_3}{25} \cdot \frac{4}{5}$.For the net electric field along the $Y$-axis to be zero,the magnitudes of these components must be equal: $\frac{K q_2}{20} \cdot \frac{4}{\sqrt{20}} = \frac{K q_3}{25} \cdot \frac{4}{5}$.Simplifying,$\frac{q_2}{20 \sqrt{20}} = \frac{q_3}{125} \Rightarrow \frac{q_2}{q_3} = \frac{20 \sqrt{20}}{125} = \frac{4 \sqrt{20}}{25} = \frac{4 \cdot 2 \sqrt{5}}{25} = \frac{8 \sqrt{5}}{25} = \frac{8}{5 \sqrt{5}}$.Comparing this with $\frac{8}{5 \sqrt{x}}$,we get $x = 5$.

If the net electric field at point $P$ along the $Y$-axis is zero,then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$,where $x = . . . . . .$

Similar Questions

The electric field due to a point charge $2q$ at a distance $r$ is $E$. Now,if charge $q$ is uniformly distributed over a thin spherical shell of radius $R$,the electric field at a distance $\frac{r}{2}$ $(r \gg R)$ from the center of the thin spherical shell is $E'=$ . . . . . . .

If the potential at the centre of a uniformly charged ring is $V_0$,then the electric field at its centre will be (assume radius $= R$):

The variation of electric field along the $Z$-axis due to a uniformly charged circular ring of radius '$a$' in the $XY$ plane is shown in the figure. The value of coordinate $M$ will be

Electric field strength due to a point charge of $5\,\mu C$ at a distance of $80\, cm$ from the charge is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module