Question:
In a space having electric field
\[
\vec{E}=A(x\hat{i}+y\hat{j})
\]
the potential at a point \((10\,m,20\,m)\) is zero, then the potential at the origin is
\[
[A=10\,Vm^{-2}]
\]
In a space having electric field
\[
\vec{E}=A(x\hat{i}+y\hat{j})
\]
the potential at a point \((10\,m,20\,m)\) is zero, then the potential at the origin is
\[ [A=10\,Vm^{-2}] \]
\[ [A=10\,Vm^{-2}] \]
Show Hint
Remember the relation between electric field and potential:
\[
\vec{E}=-\nabla V
\]
To find potential from electric field, integrate the field components with respect to their coordinates.
Updated On: Jun 24, 2026
- \(500\,V\)
- \(2000\,V\)
- \(2500\,V\)
- \(1500\,V\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Relation between electric field and potential.
The electric field is related to potential by \[ \vec{E}=-\nabla V \] Given, \[ \vec{E}=A(x\hat{i}+y\hat{j}) \] Therefore, \[ E_x=Ax \] and \[ E_y=Ay \] Using \[ E_x=-\frac{\partial V}{\partial x} \] and \[ E_y=-\frac{\partial V}{\partial y} \] we get \[ -\frac{\partial V}{\partial x}=Ax \] and \[ -\frac{\partial V}{\partial y}=Ay \]
Step 2: Find the expression for potential.
Integrating with respect to \(x\), \[ V=-\frac{A x^2}{2}+f(y) \] Similarly, integrating with respect to \(y\), \[ V=-\frac{A y^2}{2}+g(x) \] Combining both, \[ V=-\frac{A}{2}(x^2+y^2)+C \] where \(C\) is a constant.
Step 3: Use the given condition.
At the point \[ (10\,m,20\,m), \] the potential is zero.
So, \[ 0=-\frac{10}{2}(10^2+20^2)+C \] \[ 0=-5(100+400)+C \] \[ 0=-5(500)+C \] \[ 0=-2500+C \] Thus, \[ C=2500 \]
Step 4: Find the potential at the origin.
At the origin, \[ x=0,\quad y=0 \] Hence, \[ V=-\frac{10}{2}(0^2+0^2)+2500 \] \[ V=2500\,V \]
Step 5: Final conclusion.
Therefore, the potential at the origin is \[ \boxed{2500\,V} \]
The electric field is related to potential by \[ \vec{E}=-\nabla V \] Given, \[ \vec{E}=A(x\hat{i}+y\hat{j}) \] Therefore, \[ E_x=Ax \] and \[ E_y=Ay \] Using \[ E_x=-\frac{\partial V}{\partial x} \] and \[ E_y=-\frac{\partial V}{\partial y} \] we get \[ -\frac{\partial V}{\partial x}=Ax \] and \[ -\frac{\partial V}{\partial y}=Ay \]
Step 2: Find the expression for potential.
Integrating with respect to \(x\), \[ V=-\frac{A x^2}{2}+f(y) \] Similarly, integrating with respect to \(y\), \[ V=-\frac{A y^2}{2}+g(x) \] Combining both, \[ V=-\frac{A}{2}(x^2+y^2)+C \] where \(C\) is a constant.
Step 3: Use the given condition.
At the point \[ (10\,m,20\,m), \] the potential is zero.
So, \[ 0=-\frac{10}{2}(10^2+20^2)+C \] \[ 0=-5(100+400)+C \] \[ 0=-5(500)+C \] \[ 0=-2500+C \] Thus, \[ C=2500 \]
Step 4: Find the potential at the origin.
At the origin, \[ x=0,\quad y=0 \] Hence, \[ V=-\frac{10}{2}(0^2+0^2)+2500 \] \[ V=2500\,V \]
Step 5: Final conclusion.
Therefore, the potential at the origin is \[ \boxed{2500\,V} \]
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