The figure shows the electric field lines of | vedclass

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

Question

(C) For a positive charge $+q$,the electric potential at a distance $r$ is given by $V = \frac{kq}{r}$. Since $V \propto \frac{1}{r}$,the potential is

Vedclass · Accepted Answer

$+Ve, +Ve$

Vedclass · Answer

(C) For a positive charge $+q$,the electric potential at a distance $r$ is given by $V = \frac{kq}{r}$. Since $V \propto \frac{1}{r}$,the potential is higher closer to the charge.For the positive charge,point $Q$ is closer to the charge than point $P$,so $r_Q  V_P$,therefore $V_Q - V_P > 0$ (positive).For a negative charge $-q$,the electric potential is given by $V = \frac{k(-q)}{r} = -\frac{kq}{r}$. Here,the potential is more negative (lower) closer to the charge.For the negative charge,point $A$ is closer to the charge than point $B$,so $r_A  0$ (positive).Thus,both potential differences are positive.

The figure shows the electric field lines of a positive and a negative charge,respectively. Determine the sign of the potential difference $V_Q - V_P$ and $V_B - V_A$.

Similar Questions

The figure shows a positively charged infinite wire. $A$ particle of charge $q = 2 \, C$ moves from point $A$ to $B$ with constant speed. (Given linear charge density on the wire is $\lambda = 4 \pi \varepsilon_0$)

Three particles,each having a charge of $10\,\mu C$,are placed at the corners of an equilateral triangle of side $10\,cm$. The electrostatic potential energy of the system is.....$J$ (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$).

Calculate the electrostatic potential energy of a system of two charges $7\ \mu C$ and $-2\ \mu C$ placed at points $(-9\ cm, 0, 0)$ and $(9\ cm, 0, 0)$ respectively (in the absence of an external field).

$A$ proton is accelerated through $50,000 \, V$. Its energy will increase by

$A$ particle $A$ has charge $+q$ and particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference $V$,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module