Write the characteristics of equipotential surface.
Write the characteristics of equipotential surface.
$(1)$ Equipotential surface offer an alternative usual picture in addition to the picture of electric field lines around a charge configuration.
These surfaces are near in strong electric field and they are far in weak electric field.
$(2)$ No work is done in moving a test charge over an equipotential surface.
$(3)$ Electric field is always normal to the equipotential surface at every point.
$(4)$ Two equipotential surfaces can never intersect.
Similar Questions
This question has Statement $-1$ and Statement $-2$ Of the four choices given after the Statements, choose the one that best describes the two Statements
Statement $1$ : No work is required to be done to move a test charge between any two points on an equipotential surface
Statement $2$ : Electric lines of force at the equipotential surfaces are mutually perpendicular to each other
This question has Statement $-1$ and Statement $-2$ Of the four choices given after the Statements, choose the one that best describes the two Statements
Statement $1$ : No work is required to be done to move a test charge between any two points on an equipotential surface
Statement $2$ : Electric lines of force at the equipotential surfaces are mutually perpendicular to each other
- [JEE MAIN 2013]
A uniform electric field pointing in positive $x$-direction exists in a region. Let $A$ be the origin, $B$ be the point on the $x$-axis at $x = + 1$ $cm$ and $C$ be the point on the $y$-axis at $y = + 1\,cm$. Then the potentials at the points $A$, $B$ and $C$ satisfy
A uniform electric field pointing in positive $x$-direction exists in a region. Let $A$ be the origin, $B$ be the point on the $x$-axis at $x = + 1$ $cm$ and $C$ be the point on the $y$-axis at $y = + 1\,cm$. Then the potentials at the points $A$, $B$ and $C$ satisfy
- [IIT 2001]
Assertion : Two equipotential surfaces cannot cut each other.
Reason : Two equipotential surfaces are parallel to each other.
Assertion : Two equipotential surfaces cannot cut each other.
Reason : Two equipotential surfaces are parallel to each other.
- [AIIMS 2011]
A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}$ and $\frac{{{V_0}}}{4}$ have rasius $R_1,R_2,R_3$ and $R_4$ respectively. Then
A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}$ and $\frac{{{V_0}}}{4}$ have rasius $R_1,R_2,R_3$ and $R_4$ respectively. Then
- [JEE MAIN 2015]
Assertion $(A):$ A spherical equipotential surface is not possible for a point charge.
Reason $(R):$ A spherical equipotential surface is possible inside a spherical capacitor.
Assertion $(A):$ A spherical equipotential surface is not possible for a point charge.
Reason $(R):$ A spherical equipotential surface is possible inside a spherical capacitor.
- [AIIMS 2015]