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4 years ago

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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?

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:

a)

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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

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

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

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

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

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

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Thus;

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