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Bess [88]
3 years ago
13

3. A -4.00-uC charge lies 20.0cm to the right of a 2.00-uC charge on the x axis. What is the force on the 2.00-uC charge?

Physics
1 answer:
Sladkaya [172]3 years ago
3 0

<u>Answer:</u>

The force on the 2.00-uC charge is 36 \times 10^{10} \mathrm{N}

<u>Explanation:</u>

We know that force between two charges is given by Coulomb’s law,

\mathrm{F}=\mathrm{k} \frac{q 1 q 2}{r * r}

Where k = Coulomb’s constant =

9.0 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}

And q1 and q2 are the charges given to be = -4.00-uC and 2.00-uC charges

And r = distance between the charges = 20 cm = 0.2 m  

Substituting the given values in the formula we get force applied on 2.00\ \mu C charge,

F =  36 \times 10^{10} \mathrm{N} attractive force which is the required answer.

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A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

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By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

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E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

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Answer:

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Explanation:

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