Electric field vec E between two points is constant in both magnitude and direction. Consider a path

English

Gujarati

Hindi

2. Electric Potential and Capacitance

easy

The electric field $\vec E$ between two points is constant in both magnitude and direction. Consider a path of length d at an angle $\theta = 60^o$ with respect to field lines shown in figure. The potential difference between points $1$ and $2$ is

A

$\frac {E}{d\,cos\,60^o}$

B

$Ed\,cos\,60^o$

C

$\frac {Ed}{cos\,60^o}$

D

$\frac {E}{d}\,cos\,60^o$

Solution

Points $2$ and $3$ are equipotential points. Hence, potential difference between points $1$ and $2$ is the same as that between $1$ and $3$.

$\therefore V = Ed \,cos \,60^o$

Standard 12

Physics

Share

Similar Questions

Consider three concentric metallic spheres $A, B$ and $C$ of radii $a , b, c$, respectively where $a < b < c$. $A$ and $B$ are connected, whereas $C$ is grounded. The potential of the middle sphere $B$ is raised to $V$, then the charge on the sphere $C$ is

hard

(KVPY-2012)

If a charged spherical conductor of radius $10\,cm$ has potential $V$ at a point distant $5\,cm$ from its centre, then the potential at a point distant $15\,cm$ from the centre will be

easy

Assertion : Two concentric charged shells are given. The potential difference between the shells depends on charge of inner shell.

Reason : Potential due to charge of outer shell remains same at every point inside the sphere.

medium

(AIIMS-2010)

Figure shows three circular arcs, each of radius $R$ and total charge as indicated. The net electric potential at the centre of curvature is

medium

Two point charges $-Q$ and $+Q / \sqrt{3}$ are placed in the xy-plane at the origin $(0,0)$ and a point $(2,0)$, respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V =0$ in the $xy$-plane with its center at $(b, 0)$. All lengths are measured in meters.

($1$) The value of $R$ is. . . . meter.

($2$) The value of $b$ is. . . . . .meter.

normal

(IIT-2021)