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yuradex [85]
2 years ago
11

Check the photo and try me help​

Physics
1 answer:
finlep [7]2 years ago
6 0

Answer:

<h2>Ohh your photo is blurd </h2>

Explanation:

<h2>Ohh your photo is blurd </h2>
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Also included are metric units which are dekaliter onwards.
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A capacitor is to be constructed to have a capacitance of 100uF.The area of the plates is 6.om by 0.030m and the relative permit
lara [203]

Answer:

The answer is below

Explanation:

Given that:

The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m =  0.18 m²

the relative permittivity of dielectric (εr) is 7.0

Permittivity of free space (εo) = 8.854 × 10^(-12)

capacitance of 100uF

potential difference (V) of 12V

d = separation between plate

The capacitance (C) of a capacitor is given by:

C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m

The electric field between plates is given as:

E = V /d

E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m

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3 years ago
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C
miskamm [114]
<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

\boxed{q_{B}=+2q}

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential V, which is <em>the energy per unit charge, </em>so the potential V at any point in an electric field with a test charge q_{0} at that point is:

V=\frac{U}{q_{0}}

The potential V due to a single point charge q is:

V=k\frac{q}{r}

Where k is an electric constant, q is value of point charge and r is  the distance from point charge to  where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius r. Before the sphere A and B touches we have:

V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r

When they touches each other the potential is the same, so:

V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}

Therefore:

(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}

So after A and B touch and are separated the charge on sphere B is:

\boxed{q_{B}=+2q}

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

\boxed{q_{C}=+1.5q}

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

q_{A}=q_{B}=+2q

Second:  C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here q_{A}=+2q and C carries no net charge or q_{C}=0. Also, r_{A}=r_{C}=r

V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

Applying the same concept as the previous problem when sphere touches we have:

k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}

For the principle of conservation of charge:

q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q

Finally, from the third step:

Here q_{B}=+2q \ and \ q_{C}=+q. Also, r_{B}=r_{C}=r

V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

When sphere touches we have:

k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}

For the principle of conservation of charge:

q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q

So the charge that ends up on sphere C is:

q_{C}=+1.5q

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

+4q

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

+4q

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