Question

Two$10\mathrm{cm}$diameter charged disks face each other, apart. The left disk is charged to $-50\mathrm{nC}$and the right disk is charged to$+50\mathrm{nC}$.

a. What is the electric field$\overrightarrow{\mathrm{E}}$, both magnitude and direction, at the midpoint between the two disks?

b. What is the force$\overrightarrow{\mathrm{F}}$on a$-1.0\mathrm{nC}$charge placed at the midpoint?

Short answer

a.Electric field for both magnitude and direction, at the midpoint between the two disks is$-7.6\times {10}^4\mathrm{N}/\mathrm{C}·i_x$.

b.The force on alocalid="1648626759832" $-1.0\mathrm{nC}$charge placed at the midpoint islocalid="1648626764631" $7.6\times {10}^{-5}\mathrm{N}·i_x$.

Step-by-step solution

Formula for electric field

Radius of both rings is$5\mathrm{cm}$.

Charge on left ring is$-50\mathrm{nC}$.

Charge on right ring is $+50\mathrm{nC}$.

Both rings are separated bylocalid="1648627807607" $20\mathrm{cm}$distance.

Electric field due to disc of radiuslocalid="1648627817565" $\mathrm{R}$and chargelocalid="1648627827887" $\mathrm{Q}$at distancelocalid="1648627834803" $\mathrm{z}$from center of disc is given by:

localid="1648627800863" $\mathrm{E}=\frac{\mathrm{\sigma }}{2{\mathrm{ϵ}}_{\mathrm{o}}}\left[1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right]...........\left(1\right)$

Wherelocalid="1648626947333" $\sigma$is surface charge density$\sigma =\frac{Q}{\pi R^2}$.

Calculation of electric field for contribution of the left disk  (part a)

a.

Calculate the left disk's contribution,

Because the left disc is negatively charged, the electric field from it will be in the $x$direction of the left disc.

Calculate the magnitude using equation $\left(1\right)$.

$z$is equal $0.1\mathrm{m}$is half of the distance between the discs.

${\overrightarrow{\mathrm{E}}}_{\mathrm{left}}=\frac{\mathrm{Q}}{2{\mathrm{\pi R}}^2{\mathrm{ϵ}}_{\mathrm{o}}}\left[1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right]\left(-{\mathrm{i}}_{\mathrm{x}}\right)$

$=\frac{\left(50\times {10}^{-9}\mathrm{C}\right)}{2\pi (0.05\mathrm{m}{)}^2}\left[1-\frac{0.1\mathrm{m}}{\sqrt{(0.1\mathrm{m}{)}^2+(0.05\mathrm{m}{)}^2}}\right]\left(-i_x\right)$

localid="1648627784837" $\overrightarrow{{\mathrm{E}}_{\mathrm{left}}}=-3.8\times {10}^4\mathrm{N}/\mathrm{C}·{\mathrm{i}}_{\mathrm{x}}$

Calculation of electric field for contribution of the right disk and total E→part(a) solution

Calculate the right disk's contribution.

Because the distance is the same and the right disc is positively charged, the electric field will be the same due to symmetry.

${\overrightarrow{\mathrm{E}}}_{\text{right}}=\frac{\mathrm{Q}}{2{\mathrm{\pi R}}^2{\mathrm{ϵ}}_{\mathrm{o}}}\left[1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right]\left(-{\mathrm{i}}_{\mathrm{x}}\right)$

$=\frac{\left(50\times {10}^{-9}\mathrm{C}\right)}{2\pi (0.05\mathrm{m}{)}^2}\left[1-\frac{0.1\mathrm{m}}{\sqrt{(0.1\mathrm{m}{)}^2+(0.05\mathrm{m}{)}^2}}\right]\left(-i_x\right)$

localid="1648627706869" $\overrightarrow{{\mathrm{E}}_{\mathrm{right}}}=-3.8\times {10}^4\mathrm{N}/\mathrm{C}·{\mathrm{i}}_{\mathrm{x}}$

We can calculate the entire electric field by using superposition,

localid="1648627713965" $\overrightarrow{\mathrm{E}}={\overrightarrow{\mathrm{E}}}_{\mathrm{left}}+{\overrightarrow{\mathrm{E}}}_{\text{right}}$

$=-3.8\times {10}^4\mathrm{N}/\mathrm{C}·i_x-3.8\times {10}^4\mathrm{N}/\mathrm{C}·i_x$

localid="1648627725886" $\overrightarrow{\mathrm{E}}=-7.6\times {10}^4\mathrm{N}/\mathrm{C}·{\mathrm{i}}_{\mathrm{x}}$

Calculation for force (part b)

b.

Force on charge$-1.0\mathrm{nC}$will be in$+x$direction,

Magnitude is

$\overrightarrow{\mathrm{F}}=\mathrm{q}\overrightarrow{\mathrm{E}}$

$=\left(-1.0\times {10}^{-9}\mathrm{C}\right)\left(-7.6\times {10}^4\mathrm{N}/\mathrm{C}·i_x\right)$

localid="1648627754935" $\overrightarrow{\mathrm{F}}=7.6\times {10}^{-5}\mathrm{N}·{\mathrm{i}}_{\mathrm{x}}$