Question

The electric potential V in a region of space is given by $\mathbf{V}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{=}\mathbf{A}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}$where A is a constant. (a) Derive an expression for the electric field E at any point in this region. (b) The work done by the field when a $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\mu C}$test charge moves from the point $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{m}\mathbf{)}$ to the origin is measured to be 6.00 x 10-5 J. Determine A. (c) Determine the electric field at the point role="math" localid="1664863201866" $\mathbf{(}\mathbf{𝟘}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{m}\mathbf{)}$(d) Show that in every plane parallel to the xz-plane the equipotential contours are circles. (e) What is the radius of the equipotential contour corresponding to role="math" localid="1664863238356" $\mathrm{V}=1280\mathrm{V}$and $\mathrm{y}=2.00\mathrm{m}?$

Short answer

(a) Expression for the electric field E at any point in this region is$\mathrm{E}⃗=-\mathrm{A}(2\mathrm{xi}-6\mathrm{yj}+2\mathrm{zk})$

(b) The value of A is 640 V/m2

(c) The electric field at the point (0, 0, 0.250 m) is

(d) The equipotential surfaces are circles.

(e) The radius of the equipotential contour corresponding to V = 1280 V and y = 2.00 m is 3.74 m.

Step-by-step solution

Step 1:

The electric field E can be obtained if the potential V' is known. This is because the electric field is the negative of the gradient of the potential, this is

$$\begin{gathered} \mathrm{E}⃗=-\nabla \mathrm{V} \\ \mathrm{E}⃗(\mathrm{x},\mathrm{y},\mathrm{z})=-\nabla \mathrm{V}(\mathrm{x},\mathrm{y},\mathrm{z}) \\ \mathrm{V}(\mathrm{x},\mathrm{y},\mathrm{z})=\mathrm{A}({\mathrm{x}}^2-3{\mathrm{y}}^2+{\mathrm{z}}^2) \end{gathered}$$

Differentiate the given equation

$$\begin{gathered} \stackrel{r}{E}=-(î\frac{\partial V}{\partial x}+ĵ\frac{\partial V}{\partial y}+k̂\frac{\partial V}{\partial z} \\ \stackrel{r}{E}=-(2Axi-6Ayj+2Azk) \\ \stackrel{r}{E}⃗=-A(2xi-6yj+2zk) \end{gathered}$$

Expression for the electric field E at any point in this region is $\stackrel{r}{E}⃗=-A(2xi-6yj+2zk)$.

Step 2:

The electric field applies a work done ${\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}$= 6.00 x 10-5 J on a charge q =1.50 μC, where q moves from point (a) = 0,0,0.25 m to the origin to point (b) = 0,0,0.

The relation between electric field and work done is given by:

$$\begin{gathered} \frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}\mathrm{ExdI} \\ \frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}-\mathrm{A}\left(2\mathrm{xi}-6\mathrm{yj}+2\mathrm{Azk}\right)\mathrm{dI} \end{gathered}$$

The charge moves only in z-direction so the electric field in x and y directions is 0 and the equation will be in the z direction only:

localid="1664864309909" $$\begin{gathered} \frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}}=-{\int }_{\mathrm{b}}^{\mathrm{a}}-2\mathrm{Azkdz} \\ \frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}}=-{\int }_{\mathrm{b}}^{0.25}-2\mathrm{Azkdz} \\ \frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}}=\mathrm{A}{{\left[{\mathrm{z}}^2\right]}_0}^{0.25} \\ \frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}}=\mathrm{A}{(0.25\mathrm{m})}^2 \\ \mathrm{A}=\frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}{(0.25\mathrm{m})}^2} \end{gathered}$$

Substitute values to find A

$$\begin{gathered} \mathrm{A}=\frac{{\mathrm{W}}_{\mathrm{a}\to \mathrm{b}}}{\mathrm{q}{(0.25\mathrm{m})}^2} \\ =\frac{6.00\times {10}^{-6}\mathrm{J}}{\left(1.5\times {10}^{-6}\mathrm{C}\right){\left(0.25\mathrm{m}\right)}^2} \\ =640\mathrm{J}/\mathrm{C}.{\mathrm{m}}^2 \\ =640\mathrm{V}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the value of A is 640 V/m2

Step 3:

The electric field at the point (0,0,0.25m) is given by:

$$\begin{gathered} \mathrm{E}(0,0,0.250\mathrm{m})=-\mathrm{A}\left(2\right(0\widehat{)\mathrm{i}}-6\left(0\right)\hat{\mathrm{j}}+2(0.250)\hat{\mathrm{k}} \\ =-2\mathrm{A}(0.250)\hat{\mathrm{k}} \\ =-2(640\mathrm{V}/{\mathrm{m}}^2)(0.250)\hat{\mathrm{k}} \\ =-320\mathrm{V}/{\mathrm{m}}^{}\hat{\mathrm{k}} \end{gathered}$$

Therefore, the electric field at the point (0, 0, 0.250 m) is

Step 4:

The potential at xz planes has no change in the y direction therefore the term 3Ay2 for plane xz planes will be constant B and the potential at these planes will be:

$\begin{array}{l}\mathrm{V}={\mathrm{Ax}}^2+{\mathrm{Az}}^2-\mathrm{B}\\ {\mathrm{Ax}}^2+{\mathrm{Az}}^2=\mathrm{V}+\mathrm{B}\\ {\mathrm{x}}^{\mathrm{x}}+{\mathrm{z}}^2=\frac{\mathrm{V}+\mathrm{B}}{\mathrm{A}}\\ {\mathrm{x}}^{\mathrm{x}}+{\mathrm{z}}^2={\mathrm{R}}^{\mathrm{a}}\end{array}$

Where $\mathrm{R}=\sqrt{\frac{\mathrm{V}+\mathrm{B}}{\mathrm{A}}}$

Therefore, the equipotential surfaces are circles.

Step 5:

Radius R of the equipotential contour when V=12780 V and y=2.00 m given by:

$$\begin{gathered} \frac{\mathrm{V}+\mathrm{B}}{\mathrm{A}}={\mathrm{R}}^2 \\ \frac{\mathrm{V}+3{\mathrm{Ay}}^2}{\mathrm{A}}={\mathrm{R}}^2 \\ \mathrm{R}=\sqrt{\frac{\mathrm{V}+3{\mathrm{Ay}}^2}{\mathrm{A}}} \end{gathered}$$

Substitute values to get R:

$$\begin{gathered} \mathrm{R}=\sqrt{\frac{\mathrm{V}+3{\mathrm{Ay}}^2}{\mathrm{A}}} \\ =\sqrt{\frac{1280\mathrm{V}+3(640\mathrm{V}/{\mathrm{m}}^2){(2.00\mathrm{m})}^2}{640\mathrm{v}/{\mathrm{m}}^2}} \\ =3.74\mathrm{m} \end{gathered}$$

Thus, the radius of the equipotential contour corresponding to V = 1280 V and y = 2.00 m is 3.74 m.