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RideAnS [48]
3 years ago
11

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st

rength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Physics
1 answer:
Black_prince [1.1K]3 years ago
5 0

Answer:

Part A:

E_{midpoint}=0

Part B:

E_{center}=2711.7558 N/C

Explanation:

Part A:

Formula of Electric Field Strength:

E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}

Where:

x is the distance from the ring

R is the radius of the ring

\epsilon is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C

Electric Field Strength at center of left ring is same as that of right ring.

E_{center}=2711.7558 N/C

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