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What will happen to the electrostatic force between the two charges if a metal slab is placed between them?
Wiki User
∙ 11y ago
force between two charged particles always depends on their charges , distance between them and the nature of material enclosed in between them (shortly permittivity ).
permitivity is defined as the resistance encountered when forming an electric field in the medium, in other words it is defined as how much flux is generated in the medium , this flux generated would be more if the material is a metal.So, metals have high permitivity when compared to non metals or free space .
as force is inversely proportional to permitivity , metals increase permitivitty so the net force between the two charges is reduced
Wiki User
∙ 11y ago
Wiki User
∙ 13y agoThe force will be decreased by a factor of K, the dielectric constant of that dielectric sheet.
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What will happen to the objects with the same charges that are placed near to each other?
A repulsive force would exist as like charges always repel each other.
What would happen if two objects were both negatively charged?
As we know that like charges repel each other therefore if two negative charges are placed they will repel each other.
Is it possible for electrostatic lines to intersect?
No. Lines of the electrostatic field don't intersect. A 'line' of the electrostatic field is an imaginary thing that shows the force on a tiny 'test charge' placed at any point. If two 'lines' intersected, it would mean that a tiny test charge at that point would feel a force in two different directions, and would have a choice of which way to go. But that doesn't happen ... the force at any point in the field is in a single, definite direction.
Two electrostatic force between charges of 200 micro coulomb and 500 micro coulomb placed in free space in 5 gf find the distance between the two charges?
We have given that : Two charges first coloumb = 200 micro coloumb/ 200*10-6 ( there 6 is power ,1 micro coloumb = 10 di power -6/ 10-6) Second coloumb = 500 micro coloumb/500*10-6. We have given that Electrostatic Force = 5 gf ( 5 gf = 0.05 N) We need to calculate distance (r). We know that F = kq1q2/r square. R square = 9*10di power 9*200 microC*500 microC/0.05N Calculate it you will get r square = 18000mitre & r = 134.16 which is approx to 135.5 m. We know that k = 9*10di power 9.
What is the coulombs force between two point charges placed in different mediums?
In different mediums. hmm. Great question! It is related to the relative e0 constant for a particular medium. Research that.
What happens to force between two charged bodies when a glass plate is introduced in between them?
Electrostatic or magnetic charges ? the answers are different.If electrostatic charges, I GUESS that an antistatic-coated glass plate would not affect the charges at all.If it were left to accumulate charge, an insulating glass plate placed between the two charges would assume a potential between that of the two bodies. [Assuming it were free to accumulate a charge.]If magnetic charges, the glass plate would have no effect. [But the 'glass plate' equivalent would be a soft iron sheet, or a wire mesh screen of soft iron. ...Continue with your analysis of the analogy.
What is the electrostatic force of attraction between two charges of 1C and 2C placed at 5 m distance?
F=(k*1*2)/25 where k=9*10^9 in SI units
What is the force between two charge particle when medium is placed between them?
electrostatic force, which acts between two charged particles placed certain distance apart in a medium.
What is coulombs law?
Coulomb's Law states that the magnitude of the electrostatic force between two point electric charges is directly proportional to the product of the magnitudes of each charge and inversely proportional to the square of the distance between the charges. A link is provided to the Wikipedia article.
What will happen to the objects with the same charges that are placed near to each other?
A repulsive force would exist as like charges always repel each other.
Is there electrostatic dissipative mats that can be placed on an electrostatic dissipative floor?
Yes, 'Bondline' sales different types of antistatic flooring.
When a metallic sheet is placed between two charges then what will be the effect on force?
remains un changed No fancy explanation though
Why does coulomb force between two point charges depends upon dielectric constant of the intervening medium?
It is because when a dielectric is placed between the charges , the dielectric gets polarized and the net electric field between the two charges decreases, hence force = charge x electric field also decreases. john
What would happen if two objects were both negatively charged?
As we know that like charges repel each other therefore if two negative charges are placed they will repel each other.
Is it possible for electrostatic lines to intersect?
No. Lines of the electrostatic field don't intersect. A 'line' of the electrostatic field is an imaginary thing that shows the force on a tiny 'test charge' placed at any point. If two 'lines' intersected, it would mean that a tiny test charge at that point would feel a force in two different directions, and would have a choice of which way to go. But that doesn't happen ... the force at any point in the field is in a single, definite direction.
The equation of coulomb's law?
developed in the 1780s by French physicist Charles Augustin de Coulomb The magnitude of the electrostatic force between two point electric charges is directly proportional to the product of the magnitudes of each charge and inversely proportional to the square of the distance between the charges. F = k q1q2 \ d2 As F is electric force k is coulomb's constant = 9*10^9 Nm^2\C^2 q1 and q2 are the charges measured in coulombs d is the distance between them measured in meters
Two electrostatic force between charges of 200 micro coulomb and 500 micro coulomb placed in free space in 5 gf find the distance between the two charges?
We have given that : Two charges first coloumb = 200 micro coloumb/ 200*10-6 ( there 6 is power ,1 micro coloumb = 10 di power -6/ 10-6) Second coloumb = 500 micro coloumb/500*10-6. We have given that Electrostatic Force = 5 gf ( 5 gf = 0.05 N) We need to calculate distance (r). We know that F = kq1q2/r square. R square = 9*10di power 9*200 microC*500 microC/0.05N Calculate it you will get r square = 18000mitre & r = 134.16 which is approx to 135.5 m. We know that k = 9*10di power 9.
