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Marat540 [252]
3 years ago
5

a sphere with a radius of 8cm carries a uniform volume charge density of 1.5 find the magnitude of the electric field

Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

E = 5.65 x 10¹⁰ N/C

Explanation:

First we need to find the total charge on the sphere. So, we use the following formula for that purpose:

q = \sigma V\\

where,

q = total charge on sphere

V = Volume of Sphere = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.08\ m)^3 =  0.335\ m^3

σ = volume charge density = 1.5 C/m³

Therefore,

q = (0.335\ m^3)(1.5\ C/m^3) \\q = 0.502 C

Now, we use the following formula to find the electric field due to this charged sphere:

E = \frac{kq}{r^2}

where.

E  = Electric Field Magnitude = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

r = radius of sphere = 8 cm = 0.08 m

Therefore,

E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(0.502 C)}{(0.08\ m)^2}\\\\

<u>E = 5.65 x 10¹⁰ N/C</u>

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