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Aleks [24]
3 years ago
8

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
2 answers:
GalinKa [24]3 years ago
8 0

Answer:

a)The electric field at the mid-point is 0 N/C

b)The electric field at the center of the left ring is 2715.454 N/C

Explanation:

The explanation for this answers above is shown on the first,second,third,fourth,fifth image

lilavasa [31]3 years ago
3 0

Answer:

1. Electric field strength mid point is 0 N/C.

2. Electric field strength at the center of the left ring is 5420 N/C

Explanation:

D of Ring = 10cm =0.1 Meter

Charge on each Ring = Q1=Q2=40*10-9 C

Radius = D/2 = .1/2= 0.05m.

E= 1/(4piε) * (xQ/(x2+r2)^3/2)

Distance from the left ring t the mid point is x = 0.25/2 =0.125

1/4piε=8.99x10^9

E=1.8x10^4 N/C

Emid=Eleft-Eright

as Eleft=Eright

Emid=0 N/C

for Ecenter of Ring

x=0.25m

E=5.42x10^3 N/C

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7 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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3 years ago
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