Question

A uniform electric field exists in xy plane. The potential of points A\[\left( {2,2} \right)\], B\[\left( { - 2,2} \right)\] and C\[\left( {2,4} \right)\] are \[4{\rm{ V}}\], \[16{\rm{ V}}\], \[12{\rm{ V}}\] respectively. The electric field is
A. \[\left( {4\hat i + 5\hat j} \right){\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\]
B. \[\left( {3\hat i + 4\hat j} \right){\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\]
C. \[ - \left( {3\hat i + 4\hat j} \right){\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\]
D. \[\left( {3\hat i - 4\hat j} \right){\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\]

Answer 1

Hint:Electric field in a direction is equal to the partial derivative of potential in that direction. The expression for the electric field of the given x-y plane is \[\overrightarrow E = {E_x}\hat i + {E_y}\hat j\]. Here, \[{E_x}\] and \[{E_y}\] represent the electric field in x and y-direction, respectively.

Complete step by step answer:
Let us first show all the given points on the x-y plane. Here, the value of x and y is in meters.

The electric field expression in the x-direction is equal to the negative value of the ratio of change of potential and change of length in the x-direction.
\[{E_x} = - \dfrac{{\Delta {V_x}}}{{\Delta x}}\]……(1)
From the above graph, we can find:
\[
\Delta x = 2{\rm{ m}} - \left( { - 2{\rm{ m}}} \right)\\
\Rightarrow\Delta x = 4{\rm{ m}}
\]
And,
\[
\Delta {V_x} = 4{\rm{ V}} - 16{\rm{ V}}\\
\Rightarrow\Delta {V_x} = - 12{\rm{ V}}
\]
Substitute \[4{\rm{ m}}\] for \[\Delta x\] and \[ - 12{\rm{ V}}\] for \[\Delta {V_x}\] in equation (1).
\[
{E_x} = - \dfrac{{\left( { - 12{\rm{ V}}} \right)}}{{4{\rm{ m}}}}\\
\Rightarrow{E_x} = 3{\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}
\]
The electric field expression in the y-direction is equal to the negative value of the ratio of change of potential and change of length in the y-direction.
\[{E_y} = - \dfrac{{\Delta {V_y}}}{{\Delta y}}\]……(2)
Again, using the graph we can write:
\[
\Delta y = 4{\rm{ m}} - 2{\rm{ m}}\\
\Rightarrow\Delta y = 2{\rm{ m}}
\]
And,
\[
\Delta {V_y} = 12{\rm{ V}} - 4{\rm{ V}}\\
\Rightarrow\Delta {V_y} = 8{\rm{ V}}
\]
Substitute \[2{\rm{ m}}\] for \[\Delta y\] and \[8{\rm{ V}}\] for \[\Delta {V_y}\] in equation (2).
\[{E_y} = - \dfrac{{8{\rm{ V}}}}{{2{\rm{ m}}}}\\
\Rightarrow{E_y} = - 4{\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}
\]
Let us now write the expression of the magnetic field in vector form.
\[\overrightarrow E = {E_x}\hat i + {E_y}\hat j\]
Here \[\hat i\]and \[\hat j\] are unit vectors in x and y direction, respectively.
Substitute \[3{\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\] for \[{E_x}\] and \[ - 4{\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\] for \[{E_y}\] in the above expression of the electric field.
\[
\overrightarrow E = \left( {3{\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}} \right)\hat i + \left( { - 4{\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}} \right)\hat j\\
\therefore\overrightarrow E = \left( {3\hat i - 4\hat j} \right){\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}
\]

Therefore, the expression for the electric field is \[\left( {3\hat i - 4\hat j} \right){\rm{ }}{{\rm{V}} {\left/
 {\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\] , and option (D) is correct.

Note: Alternate method: We can assume the expression for the potential of any point (x, y). Substituting the individual values of x, y, and potential in that equation and solving them, we will get unknown quantities. On substituting the unknown quantities in the expression of potential, we can obtain the general expression of the potential and electric field.