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erica [24]
4 years ago
13

A 2.0 m × 4.0 m horizontal plastic sheet has a charge of −10 μC , uniformly distributed. A tiny 4.0 μg plastic sphere is suspend

ed motionless just above the center of the sheet. What is the charge on the sphere?
Physics
1 answer:
gtnhenbr [62]4 years ago
7 0

Answer:

Explanation:

Given

Dimension of Plastic sheet is 2\times 4\ m^2

Charge on sheet Q=-10\ \mu C

Charge density \sigma =\frac{q}{A}

\sigma =\frac{-10\times 10^{-6}}{8}=1.25\times 10^{-6}\ C/m^2

Sphere has mass of  m=4\ \mug

If sphere is suspended motionless then its weight is balanced by repulsion force

Repulsive force F_r=qE

where E=Electric field due to sheet

E=\frac{q}{2\epsilon _0}

E=\frac{-10\times 10^{-6}}{2\times 8.85\times 10^{-12}}

E=5.647\times 10^{5}\ N/C

F_r=q\times 5.647\times 10^{5}

F_r=mg

q=\frac{mg}{E}

q=\frac{4\times 10^{-6}\times 9.8}{5.647\times 10^{5}}

q=6.9417\times 10^{-11}\ C

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Explanation:

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        where v₀ is the initial velocity and a is the acceleration, being t the

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3 years ago
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
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Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

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k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

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Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

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\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

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Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

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