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# Tag Archives: charge

## Flow of electrons and direction of current

We say current flow opposite to electron flow but what is current?

Electric current is defined as the rate of flow of charges, ie; the amount of charges flowing in 1 sec.

The direction of current was conventionally defined as the direction of flow of positive charges.
In a metallic conductor, the negatively charged particles (electrons) are moving and therefore the direction of current is taken as opposite to the direction of flow of electrons.
Hope the concept is clear.

## Capacitance, voltage and Potential difference

the capacity of parallel plate condenser is 5 micro Faraday. When a glass plate is placed between the plates of the conductor  its potential become 1/8th of the original value. The value of dielectric constant will be
(A)1.6
(B)5
(C)8
(D)40

The charge remains same

Use Cm=KCo
and
V reduced to 1/8 means capacitance increased 8 times since Q=CV
So, the answer is evident

## Size of a soap bubble when charged

What will happen to the size of the soap bubble if the bubble is given charges ?

Answer: When a soap bubble is charged, its size increases.

Under normal condictions, a soap bubble is in equilibrium un der two opposing forces – the force of surface tension which tries to compress it and the force due to excess pressure which tries to expand it.

When charged, the force of repulsion among the like charges will try to expand the soap bubble further resulting in an increase in size of the bubble.

## Potential Difference

“What is meant by potential difference?”

“Potential Difference between two points is the difference in Potential between the two points and is equal to the work done per unit charge ion carrying a positive test charge from one point to another against the electric field and without any acceleration”

## de Broglie Wavelength

Alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- broglie wavelength

Charge of alpha particle = 2e

Mass of alpha particle = 4 u

Charge of proton = e

mass of proton = u

The energy acquired by proton when accelerated through a pd of V,

E=eV

The momentum acquired by proton=${\sqrt{2ueV}}$

The de Broglie wavelength is given by $\lambda =\frac{h}{mv}$

Therefore, de Broglie wavelength of Proton, $\lambda _{proton}=\frac{h}{\sqrt{2ueV}}$

Similarly,$\lambda _{alpha}=\frac{h}{\sqrt{2\times 4u\times 2e\times V}}$

$\frac{\lambda _{alpha}}{\lambda _{proton}}=\frac{\sqrt{2ueV}}{\sqrt{2\times 4u\times 2e\times V}}=\frac{1}{2\sqrt{2}}$

## Electric Field of Axons!

A nerve signal is transmitted through a neuron when an excess Na+ inos suddenly enters the axon, along cylindrical part of the neuron. Axons are approximately 10.0 micrometer in diameter, and measurements show that about 5.6×10^11 Na+ ions per meter ( each of charge e+)enter during this process. Although the axon is a long cylinder, the charge doesn’t all enter everywhere at the same time. A plausible model would be a series if point charges moving along the axon. Let us look at a 0.10-mm length of the axon and model it as a point of charge.

(a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10-mm length of the axon?
(b) What electric field ( magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin?
(c) Certain sharks can respond to electric fields as weak as 1.0 microN/C. how far from this segment of axon could a shark be and still detect its electric field?

## Motion of a charge particle in an electric field – Numerical Problem

At t=0 a very small object with mass 0.400mg and charge +9.00μC is travelling at 125 m/s in the -x-direction. The charge is moving in a uniform electric field that is in the +y-direction and that has magnitude E= 895 N/C. The gravitational force on the particle can be neglected. How far is the particle from the origin at t= 7.99 ms.

Hint for solving the above problem:

displacement along X direction; x=-uxt=125 x 7.99/1000

Displacement along Y direction, y=0.5 ay t2

ay=qE/m

s=√(x2+y2)

Just substitute the values and get the answer

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