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maw [93]
3 years ago
7

At what distance along the central perpendicular axis of a uniformly charged plastic disk of radius 0.390 m is the magnitude of

the electric field equal to one-half the magnitude of the field at the center of the surface of the disk?
Physics
1 answer:
aliina [53]3 years ago
3 0
<span>Radius, the distance from the centre = 0.390
 Electric field is equal to half of the magnitude. E2 = E / 2
 Given
E1 = E2 E1 = k x Q / r^2
  E2 = (k x Q / r2^2) / 2
  Equating the both we get 2 x r^2 = r2^2
 r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
  r2 = 1.414 x 0.390 = 0.551 m</span>
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Answer:

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Explanation:

Given that,

Radius of sphere R=2.81 cm

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(a). We need to calculate the potential at a distance r = 1.60 cm

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V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr

V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr

V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}

V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})

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The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). We need to calculate the potential at a distance r = R

Using formula of  potential difference

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V_{R}=-0.0003759\ V

V_{R}=-3.759\times10^{-4}\ V

The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Hence, This is the required solution.

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3 years ago
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Answer:

Explanation:

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Answer: Gravitational force

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