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What will happen to the size of the soap bubble if the bubble is given charges ?
Asked VARSHA UDAYAKUMAR
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.
The mass of a proton is 1840 times that of an electron. it is accelerated to a potential difference of 1kV. find the amount of work done in process. (Ujjwal Sharma asked)
The work done = the KE acquired = eV=1.6 x 10-19 x 1000 = 1.6 x 10-16 J
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- An infinite number of charges,each of magnitude q,are placed along X-axis at X=1m,2m,4m,8m,16m and so on but the consecutive charges are of opposite sign starting with +q at x=1m. a point charge q’, kept at the origin ,experiences a force of what magnitude?
- four point charges,each +q,are fixed at the corners of a square of side ‘a’. another point charge q’ is placed at a height ‘h’ vertically above the centre of square,assuming the square to be in horizontal plane. what is the magnitude of force experienced by q’?
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Can you please answer the following question?
Consider the charges q,q and -q placed at the vertices of an equilateral triangle. What is the force on each charge?
I actually didn’t understand the total force on charge -q at the vertex C of the equilateral triangle. How is it equal to root3Fn where n is the unit vector along the direction bisecting the angle BCA (Why?).
Thejaswini posted the question
What are the limitations of Van de Graff Generator?
One of the limitations of Van de Graff generator is the potential to which the dome can be raised. In normal condition, discharge takes place when the potential reaches 3 x 106 V. The potential can be increased further by placing the entire system in a container filled with high-pressure gas.
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How do you find the direction of electric field on the outer or inner surface of 2 parallel plate of a capacitor ?
The direction of electric field is from the positive plate to negative plate.
The electric field outside the plates is zero as the electric field due to each plate (E=Q/2ε0A) cancel out being equal in magnitude and opposite in direction.
In between the plates the two electric fields add up as they are in same direction. (From positive plate to negative plate).
The magnitude of electric field between the plates is twice the electric field due to either; i.e; E=2 x Q/2ε0A = Q/ε0A