The electric potential existing in space is \[\hspace{0.167em} V(x, y, z) = A(xy + yz + zx) .\] (a) Write the dimensional formula of A. (b) Find the expression for the electric field. (c) If A is 10 SI units, find the magnitude of the electric field at (1 m, 1 m, 1 m).
Solution
Given:
Electric potential,
\[V(x, y, z) = A(xy + yz + zx)\]
\[A = \frac{\text{ volt }}{m^2}\]
\[ \Rightarrow \left[ A \right] = \frac{\left[ {ML}^2 I^{- 1} T^{- 3} \right]}{\left[ L^2 \right]}\]
\[ \Rightarrow A = [ {MT}^{- 3} I^{- 1} ]\]
(b) Let E be the electric field.
\[dV = - \vec{E} . \vec{dr} \]
\[ \Rightarrow A(y + z)dx + A(z + x)dy + A(x + y)dz = - E(dx \hat{i} + dy \hat{j} + dz\hat{ k } )\]
\[ \Rightarrow [A(y + z) \hat{i } + A(z + x)\hat{ j } + A(x + y) \hat{ k } ] [dx\hat{ i} + dy \hat{j } + dz \hat{k } ] = - E\left[ dx \hat{ i }+ dy\hat{ j } + dz \hat{ k } \right]\]
Equating now, we get
\[\vec{E} = - A(y + z) \hat{ i } - A(z + x) \hat{ j } - A(x + y) \hat{ k }\]
(c) Given: A = 10 V/m2
\[r = (1 m, 1 m, 1 m)\]
\[ \vec{E} = - 10 (2) \hat{ i } - 10 (2) \hat{ j } - 10 (2) \hat{ k } \]
\[ = - 20 \hat{ i } - 20 \hat{ j } - 20 \hat{ k }\]
Magnitude of electric field,
\[\left| E \right| = \sqrt{{20}^2 + {20}^2 + {20}^2}\]
\[ = \sqrt{1200} = 34 . 64 = 35\] N/C
