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astra-53 [7]
2 years ago
7

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

10 to the power -5 C? What is the total energy stored in it?
Physics
1 answer:
Maurinko [17]2 years ago
6 0

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

Learn more about capacitance here:

brainly.com/question/14883923

#SPJ1

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OlgaM077 [116]

Answer:

The boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Explanation:

The boat only moves because the centre of mass changes a bit if the two people on opposite ends of the boat exchange seats.

The boat moves a distance of the change in centre of mass

Noting that the weight of the boat acts at the centre of the boat at 1.45m from both ends.

For convention, we call the original position of the 59 kg person as x=0

This means,

59 kg person is at x = 0 m

88 kg of the boat acts at x = 1.45 m from the end of the 59 kg person.

78 kg person is at x = 2.90 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

For the initial setup,

X = [(59×0) + (88×1.45) + (78×2.90)]/(59+88+78)

X = (353.8/225)

X = 1.572 m

(Don't forget that this is 1.572 m from the end we designated x=0 m)

When the people exchange positions,

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88 kg of the boat still acts at the centre of the boat at x = 1.45 m

And 78 kg person is now at the end of the boat with x = 0 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

X = [(59×2.90) + (88×1.45) + (78×0)]/(59+88+78)

X = (298.7/225)

X = 1.328 m

(This is 1.328 m from the end we designated x=0 m from the start)

So, the centre of mass moves a distance of (1.572 - 1.328) towards the end of the boat we designated x=0 m from the start.

Hence, the boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Hope this Helps!!!

6 0
3 years ago
An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect
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Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is  B= 0.0048 T

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

                             \Delta V q = \frac{1}{2}mv^2

      Making v the subject

                             v = \sqrt{[\frac{2 \Delta V * q }{m}] }

 Where m is the mass of electron

              v is the velocity of electron

              q charge on electron

               \Delta V is the potential difference  

Substituting values

         v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }f

            = 4.5556 *10^ {7} m/s

For the electron to move in a circular path the magnetic force[F = B q v] must be equal to the centripetal force[\frac{mv^2}{r}] and this is mathematically represented as

                  Bqv = \frac{mv^2}{r}

making B the subject

                B = \frac{mv}{rq}

r is the radius with a value = 5.4cm = = \frac{5.4}{100} = 5.4*10^{-2} m

Substituting values

                B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}

                     = 0.0048 T

                 

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