Helppp! What do we call the low of electrons through conductive material?

A parallel plate air capacitor has a capacitance of 10 to the power -9. What potential difference is required for a charge of 5×

Maurinko [17]

The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules

<h3>What is the energy in a capacitor?</h3>

The energy stored in a capacitor is an electrostatic potential energy.

It is related to the charge(Q) and voltage (V) between the capacitor plates.

It is represented as 'U'.

<h3>How to determine the potential difference</h3>

Formula:

Potential difference, V is the ratio of the charge to the capacitance of a capacitor.

It is calculated using:

V = Q ÷ C

Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9

Substitute the values into the equation

Potential difference, V = 5 × 10∧-5 ÷ 10∧-9 = 5 × 10∧4 volts

<h3>How to determine the energy stored</h3>

Formula:

Energy, U = 1 ÷ 2 (QV)

Where Q= charge and V = potential difference across the capacitor

Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)

= 0.5 × 25 × 10∧-1

= 0.5 × 2.5

= 1. 25 Joules

Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules

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6 0

2 years ago

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

sergij07 [2.7K]

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

$V = iR$

$V = (2 mA)(10 kohm)$

$V = 20 volts$

so voltage across the capacitor + voltage across resistor = V

$V_c + 20 = 50$

$V_c = 30 V$

Now we know that

$U = \frac{q^2}{2C}$

here rate of change in energy of the capacitor is given as

$\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}$

$\frac{dU}{dt} = (30)(2 mA)$

$\frac{dU}{dt} = 0.06 W$

3 0

3 years ago

How do the current and coltage readings compare to before the wires were switched. Explain how to interpret them physically?

Blababa [14]

The potential difference between two locations in an electric circuit is measured using a voltmeter.
If the electricity passes through the voltmeter it shows deflection.

<h3>What is the purpose of a voltmeter?</h3>

A voltage meter, usually referred to as a voltmeter, is a device that measures the voltage, or potential difference, between two points in an electrical or electronic circuit.
volts is the unit of voltmeter(volts, millivolts, kilovolts)

<h3>What is the explanation for the link between current and voltage?</h3>

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

<h3>What is ohm's law in circuit?</h3>

V = IR, where V is voltage, I is current, and R is resistance, is known as Ohm's Law.
If you know the voltage of the battery in the circuit and how much resistance is in the circuit, you may use Ohm's Law to identify properties of a circuit, such as how much current is flowing through it.

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7 0

2 years ago

Which of the following statements is true about the scientific process?

KiRa [710]

The best and most correct answer among the choices provided by your question is the third choice or letter C.
<span>The statement "Your hypothesis must be testable." is true about the scientific process.

</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

3 years ago

what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?

zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

5 0

3 years ago