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Anuta_ua [19.1K]
3 years ago
7

A solid conducting sphere with radius 0.75 m carries a net charge of 0.13 nC. What is the magnitude of the electric field at a p

oint located 0.50 from the sphere's center 0.25 beneath the sphere's surface)?
Physics
2 answers:
ludmilkaskok [199]3 years ago
5 0

Answer:

Magnitude of Electric field E at at point 0.50m which is within the sphere is Zero( i.e E = 0)

Explanation:

It is understand from Guass' law, the electric field in a region enclosed by a conducting sphere is Zero.

It is given that the radius of the sphere is 0.75m, therefore, a point located at 0.50m from the sphere centre 0.25m before the sphere surface still falls inside the sphere, therefore making the electric field at that point zero in magnitude.

Delicious77 [7]3 years ago
3 0

Answer:

Explanation:

given that

Radius =0.75m

Cnet=0.13nC

a. Electric field inside the sphere located 0.5m from the center of the sphere.

The electric field located inside the sphere is zero.

b. The electric field located 0.25m beneath the sphere.

Since the radius is 0.75m

Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m

Then

E=kq/r2

K=9e9Nm2/C2

q=0.13e-9C

r=1m

Then,

E= 9e9×0.13e-9/1^2

E=1.17N/C. Q.E.D

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Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

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Where;

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We are given;

m = 1.81 × 10^(−3) kg

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V = (3.00 × 10⁴ m/s) ȷ^.

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Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

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a = -0.33 m/s² k^

7 0
3 years ago
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bearhunter [10]

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