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Svetach [21]
3 years ago
12

Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t

he same nail in a ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes a 13.0 ∘ angle with the vertical. How many surplus electrons are on each sphere?
Physics
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

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Answer:

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Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

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V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

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Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0
3 years ago
A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha
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Answer:

0.358Kg

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Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

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