91Ó°ÊÓ

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" $\mathrm{E}=\frac{\mathrm{σ}}{2{\mathrm{Ï\mu }}_{\mathrm{o}}}\left[1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right]...........\left(1\right)$

Wherelocalid="1648626947333" $\mathrm{σ}$is surface charge density$\mathrm{σ}=\frac{Q}{\mathrm{π}{R}^2}$.

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.

${\stackrel{→}{\mathrm{E}}}_{\mathrm{left}}=\frac{\mathrm{Q}}{2{\mathrm{Ï€¸é}}^2{\mathrm{Ï\mu }}_{\mathrm{o}}}\left[1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right]\left(-{\mathrm{i}}_{\mathrm{x}}\right)$

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

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

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.

${\stackrel{→}{\mathrm{E}}}_{\text{right}}=\frac{\mathrm{Q}}{2{\mathrm{Ï€¸é}}^2{\mathrm{Ï\mu }}_{\mathrm{o}}}\left[1-\frac{\mathrm{z}}{\sqrt{{\mathrm{z}}^2+{\mathrm{R}}^2}}\right]\left(-{\mathrm{i}}_{\mathrm{x}}\right)$

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

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

We can calculate the entire electric field by using superposition,

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

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

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

b.

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

Magnitude is

$\stackrel{→}{\mathrm{F}}=\mathrm{q}\stackrel{→}{\mathrm{E}}$

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

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