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Calculate the capacitance of a system that stores 9.4 x 10-10 C of charge at

pantera1 [17]

The capacitance of the system is $1.9\cdot 10^{-11}F$

Explanation:

The capacitance of a capacitor is given by the following equation:

$C=\frac{Q}{V}$

where

C is the capacitance

Q is the charge stored in the capacitor

V is the potential difference across the capacitor

In this problem, we have:

$Q=9.4\cdot 10^{-10}C$ is the charge stored

$V=50.0 V$ is the potential difference across it

Substituting into the equation, we find the capacitance:

$C=\frac{9.4\cdot 10^{-10}}{50.0}=1.9\cdot 10^{-11}F$

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7 0

3 years ago

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You toss a ball in the air. what is the work done by gravity as the ball goes up?

jolli1 [7]

Answer: Gravity slows the ball down as it goes up and eventually stops it from going up and starts to pull it back down to earth.

Explanation:

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2 years ago

What is the difference between a wedge and a screw?

Anvisha [2.4K]

<span>A wedge is a double sided screw basically. Therefore, the difference is the number of sides.</span>

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3 years ago

Define heat capacity of a substance.<br>Write the S.I unit of heat capacity.

creativ13 [48]

Define heat capacity of a substance:

The heat capacity of a substance can be defined as the amount of heat required to change its temperature by one degree.

Write the S.I unit of heat capacity:

Joule per Kelvin

$\: \: \:$

Joule per Celsius

$\: \: \:$

-,-

6 0

1 year ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago