A point charge q is situated at a distance d | vedclass

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

Question

$\mathrm{F}=\int_{\mathrm{d}}^{\mathrm{d}+\mathrm{L}} \frac{\mathrm{kq} \mathrm{dq}}{\mathrm{x}^{2}}=\int_{\mathrm{d}}^{\mathrm{d}+\mathrm{L}} \frac{\

Vedclass · Accepted Answer

$\frac{1}{{4\pi \,{ \in _0}}}\frac{{qQ}}{{d\left( {d + L} \right)}}$

Vedclass · Answer

$\mathrm{F}=\int_{\mathrm{d}}^{\mathrm{d}+\mathrm{L}} \frac{\mathrm{kq} \mathrm{dq}}{\mathrm{x}^{2}}=\int_{\mathrm{d}}^{\mathrm{d}+\mathrm{L}} \frac{\mathrm{kq}\left(\frac{\mathrm{Q}}{\mathrm{L}} \mathrm{dx}\right)}{\mathrm{x}^{2}}=\frac{\mathrm{KqQ}}{\mathrm{L}} \int_{\mathrm{d}}^{d+\mathrm{L}} \frac{1}{\mathrm{x}^{2}} \mathrm{dx}$

$\mathrm{F}=\frac{\mathrm{KqQ}}{\mathrm{d}(\mathrm{d}+\mathrm{L})}=\frac{1}{4 \pi \varepsilon_{\mathrm{o}}} \frac{\mathrm{qQ}}{\mathrm{d}(\mathrm{d}+\mathrm{L})}$

A point charge $q$ is situated at a distance $d$ from one end of a thin non - conducting rod of length $L$ having a charge $Q$ (uniformly distributed along its length) as shown in fig.Then the magnitude of electric force between them is

Similar Questions

A wheel having mass $m$ has charges $+q $ and $-q$ on diametrically opposite points. It remains in equilibrium on a rough inclined plane in the presence of uniform vertical electric field $E =$

What is the equivalent capacitance of the system of capacitors between $A$ and $B$ :-

Two charges $+q$ and $-3q$ are placed on $x-$ axis separated by a distance $d$. ($-3q$ is right of $q$) Where should a third charge $2q$ be placed such that it will not experience any force ?

Find flux related to shaded face $BCGF$