(A) The electric potential at point $A$ is $V_A$.Since the electric field $E$ is uniform and directed towards the right,the potential decreases in the

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(A) The electric potential at point $A$ is $V_A$.Since the electric field $E$ is uniform and directed towards the right,the potential decreases in the direction of the electric field.The horizontal distance between $A$ and $B$ is $x$. Therefore,the potential difference between $A$ and $B$ is $\Delta V = V_B - V_A = -E \cdot x$.Thus,the potential at $B$ is $V_B = V_A - Ex$.The potential energy $U$ of a charge $q$ at a point with potential $V$ is given by $U = qV$.Therefore,the potential energy of the charge at point $B$ is $U_B = q V_B = q(V_A - Ex)$.

Class 12 Physics Electric Potential and Capacitance Electric potential

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$A$ uniform electric field vector $E$ exists along the horizontal direction as shown. The electric potential at $A$ is $V_A$. $A$ small point charge $q$ is slowly moved from $A$ to $B$ along the curved path as shown. The potential energy of the charge when it is at point $B$ is

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A
$q(V_A - Ex)$
B
$q(V_A + Ex)$
C
$q(Ex - V_A)$
D
$qEx$

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Electric Potential and Capacitance

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Electric potential

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Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a << r$) is: ($\varepsilon_0$ is the permittivity of free space)

DifficultAP EAMCET 2007

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Four identical charges $+50\,\mu C$ each are placed,one at each corner of a square of side $2\,m$. How much external energy is required to bring another charge of $+50\,\mu C$ from infinity to the centre of the square? (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\frac{N\cdot m^2}{C^2}$)

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An electric charge $10^{-6} \mu C$ is placed at the origin $(0,0) \text{ m}$ of an $X-Y$ coordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3}) \text{ m}$ and $(\sqrt{6}, 0) \text{ m}$ respectively. The potential difference between the points $P$ and $Q$ will be:

DifficultJEE MAIN 2024

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The electric potential inside a conducting sphere:

Easy

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Two conducting spheres of radii $r_1$ and $r_2$ have the same electric field intensity near their surfaces. The ratio of their electric potentials is

Medium

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$A$ uniform electric field vector $E$ exists along the horizontal direction as shown. The electric potential at $A$ is $V_A$. $A$ small point charge $q$ is slowly moved from $A$ to $B$ along the curved path as shown. The potential energy of the charge when it is at point $B$ is

Similar Questions

Along the $X$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at $x=a+r$ (if $a << r$) is: ($\varepsilon_0$ is the permittivity of free space)

Four identical charges $+50\,\mu C$ each are placed,one at each corner of a square of side $2\,m$. How much external energy is required to bring another charge of $+50\,\mu C$ from infinity to the centre of the square? (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\frac{N\cdot m^2}{C^2}$)

An electric charge $10^{-6} \mu C$ is placed at the origin $(0,0) \text{ m}$ of an $X-Y$ coordinate system. Two points $P$ and $Q$ are situated at $(\sqrt{3}, \sqrt{3}) \text{ m}$ and $(\sqrt{6}, 0) \text{ m}$ respectively. The potential difference between the points $P$ and $Q$ will be:

The electric potential inside a conducting sphere:

Two conducting spheres of radii $r_1$ and $r_2$ have the same electric field intensity near their surfaces. The ratio of their electric potentials is

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