How do you increase capacitance in a capacitor?

1 answer:

4 0

If you want to increase the Capacitance of Parallele Plate Capacitor then increase the surface area, reduce the separation between the plate and use a dielectric material in between the plate which have higher dielectric breakdown strength.

You might be interested in

Find the current in the 12 ohm resistor.

disa [49]

Answer:

1.5 A

Explanation:

V =I.R

18 = I × 12

I = 18/12

= 3/2 = 1.5 A

3 0

2 years ago

A turntable must spin at 33.4 rpm (3.50 rad/s) to play an old-fashioned vinyl record. how much torque must the motor deliver if

Westkost [7]

In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration

For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds

α = (3.5 - 0) / 1.8
α = 1.94 rad/s²

The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002

τ = 1.94 * 0.002
τ = 0.004

The torque is 0.004

4 0

3 years ago

1. Predict whether the energy required to remove an electron from magnesium and potassium would be more or less than that requir

Inga [223]

In both cases less energy is required

But comparetively Mg require more energy than K

Let's see the electron configuration of Both

[Mg]=1s²2s²2p⁶3s²=[Ne]3s²
[K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹

K has only one valence electron so very less ionization enthalpy so less energy required

Mg has 2 so more IE hence more energy required

8 0

1 year ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

3 years ago

Can charging by friction only occur in solids

Sonja [21]

Yes you are right cuz charging by friction cant be done in fluids( liquid and gas)

6 0

2 years ago