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A uniform surface charge of σ = 2 μC/m2 , is situated at z = 2 plane. What is the value of flux density at P(1,1,1)m?
To determine the electric flux density (mathbf{D}) at point (P(1,1,1)) due to a uniform surface charge density (sigma = 2 , mu C/m^2) located at the plane (z = 2), we can refer to Gauss's law in differential form:[nabla cdot mathbf{D} = rho_v]However, to apply Gauss's law directly in this context, iRead more
To determine the electric flux density (mathbf{D}) at point (P(1,1,1)) due to a uniform surface charge density (sigma = 2 , mu C/m^2) located at the plane (z = 2), we can refer to Gauss’s law in differential form:
[
nabla cdot mathbf{D} = rho_v
]
However, to apply Gauss’s law directly in this context, it’s easier to consider its integral form, especially since we are dealing with a surface charge in a specific location:
[
oint_S mathbf{D} cdot dmathbf{A} = Q_{enc}
]
Where (oint_S mathbf{D} cdot dmathbf{A}) is the electric flux through a closed surface (S) and (Q_{enc}) is the total charge enclosed by that surface. Since the point of interest (P) is not enclosed by the charged surface and the charge distribution lies entirely in a plane at (z=2), the scenario simplifies the analysis.
For an infinite plane with surface charge density (sigma), the electric field (E) (and thus, the flux density (mathbf{D})) is constant and perpendicular to the surface. By symmetry, above and below the plane, the electric field points away or towards the plane for positive
See lessDetermine the charge that produces an electric field strength of 40 V/cm at a distance of 30cm in vacuum(in 10-8C)
To find the charge that produces an electric field strength of 40 V/cm at a distance of 30 cm in a vacuum, we can use Coulomb's law. The electric field ((E)) at a distance (r) from a point charge (q) in a vacuum is given by the formula:[E = frac{kq}{r^2}]where (k) is Coulomb's constant ((8.987 timesRead more
To find the charge that produces an electric field strength of 40 V/cm at a distance of 30 cm in a vacuum, we can use Coulomb’s law. The electric field ((E)) at a distance (r) from a point charge (q) in a vacuum is given by the formula:
[E = frac{kq}{r^2}]
where (k) is Coulomb’s constant ((8.987 times 10^9 , text{N m}^2/text{C}^2)), (q) is the charge in coulombs (C), and (r) is the distance from the charge in meters.
Given that the electric field strength (E = 40, text{V/cm} = 4000, text{V/m}) (since 1 V/m = 0.01 V/cm), and the distance (r = 30, text{cm} = 0.3, text{m}), we can rearrange the formula to solve for (q):
[q = frac{E cdot r^2}{k}]
Substitute the given values into the equation:
[q = frac{4000 cdot (0.3)^2}{8.987 times 10^9}]
[q = frac{4000 cdot 0.09}{8.987
See lessA charge of 2 X 10-7 C is acted upon by a force of 0.1N. Determine the distance to the other charge of 4.5 X 10-7 C, both the charges are in vacuum
To determine the distance between two charges in a vacuum, you would typically use Coulomb's Law, which is given by the equation:[ F = k cdot frac{{|q_1 cdot q_2|}}{{r^2}} ]Where:- (F) is the force between the charges (in Newtons, N),- (k) is Coulomb's constant ((8.987 times 10^9 , text{N} cdot textRead more
To determine the distance between two charges in a vacuum, you would typically use Coulomb’s Law, which is given by the equation:
[ F = k cdot frac{{|q_1 cdot q_2|}}{{r^2}} ]
Where:
– (F) is the force between the charges (in Newtons, N),
– (k) is Coulomb’s constant ((8.987 times 10^9 , text{N} cdot text{m}^2/text{C}^2)),
– (q_1) and (q_2) are the magnitudes of the two charges (in Coulombs, C),
– (r) is the distance between the centers of the two charges (in meters, m).
Given:
– (q_1 = 2 times 10^{-7} , text{C}),
– (q_2 = 4.5 times 10^{-7} , text{C}),
– (F = 0.1 , text{N}).
Substitute the given values into Coulomb’s Law and solve for (r):
[ 0.1 = (8.987 times 10^9) cdot frac{{|2 times 10^{-7} cdot 4.5 times 10^{-7}|}}{{r^2}} ]
First,
See lessTwo small diameter 10gm dielectric balls can slide freely on a vertical channel. Each carry a negative charge of 1μC. Find the separation between the balls if the lower ball is restrained from moving
To find the separation between the two charged dielectric balls, we can employ Coulomb's Law and the equilibrium condition due to the gravitational force acting on the upper ball.Coulomb's Law gives us the electrostatic force (F_e) between two charges:[ F_e = k_e frac{|q_1 q_2|}{r^2} ]where- (k_e =Read more
To find the separation between the two charged dielectric balls, we can employ Coulomb’s Law and the equilibrium condition due to the gravitational force acting on the upper ball.
Coulomb’s Law gives us the electrostatic force (F_e) between two charges:
[ F_e = k_e frac{|q_1 q_2|}{r^2} ]
where
– (k_e = 8.99 times 10^9 , text{N}cdottext{m}^2/text{C}^2) is the Coulomb’s constant,
– (q_1) and (q_2) are the charges of the balls, which are each (1 mu C = 1 times 10^{-6} C),
– (r) is the separation between the centers of the two balls, which we are trying to find.
The gravitational force (F_g) acting on the upper ball is given by:
[ F_g = mg ]
where
– (m = 10 , text{gm} = 0.01 , text{kg}) is the mass of the ball,
– (g = 9.8 , text{m/s}^2) is the acceleration due to gravity.
At equilibrium, the electrostatic force of repulsion between the balls is equal to the gravitational force pulling the upper ball downwards:
[ F_e = F_g
See lessFind the force between two charges when they are brought in contact and separated by 4cm apart, charges are 2nC and -1nC, in μN
To find the force between two charges, we can use Coulomb's Law, which is expressed as:[F = k cdot frac{|q_1 cdot q_2|}{r^2}]where:- (F) is the force between the charges,- (k) is Coulomb's constant ((8.987 times 10^9 N m^2/C^2)),- (q_1) and (q_2) are the magnitudes of the two charges (in this case,Read more
To find the force between two charges, we can use Coulomb’s Law, which is expressed as:
[F = k cdot frac{|q_1 cdot q_2|}{r^2}]
where:
– (F) is the force between the charges,
– (k) is Coulomb’s constant ((8.987 times 10^9 N m^2/C^2)),
– (q_1) and (q_2) are the magnitudes of the two charges (in this case, 2nC and -1nC, which are (2 times 10^{-9}C) and (-1 times 10^{-9}C) respectively),
– and (r) is the distance between the charges (4cm, which needs to be converted to meters, so (r = 0.04 m)).
Substituting the given values:
[F = 8.987 times 10^9 N m^2/C^2 cdot frac{(2 times 10^{-9} C) cdot (1 times 10^{-9} C)}{(0.04 m)^2}]
[F = 8.987 times 10^9 cdot frac{2 times 10^{-18}}{0.0016}]
[F = 8.987 times 10^9 cdot 1.
See lessFind the force of interaction between 60 stat coulomb and 37.5 stat coulomb spaced 7.5cm apart in transformer oil(εr=2.2) in 10-4 N
To calculate the force of interaction between two charges in a medium other than vacuum, we can use Coulomb's Law, modified to account for the dielectric constant of the medium ((epsilon_r)).The formula for the force between two point charges in a vacuum is given by Coulomb's law as:[F = k frac{{|q_Read more
To calculate the force of interaction between two charges in a medium other than vacuum, we can use Coulomb’s Law, modified to account for the dielectric constant of the medium ((epsilon_r)).
The formula for the force between two point charges in a vacuum is given by Coulomb’s law as:
[
F = k frac{{|q_1 q_2|}}{{r^2}}
]
where
– (F) is the force between the charges,
– (q_1) and (q_2) are the magnitudes of the two charges,
– (r) is the distance between the charges, and
– (k) is Coulomb’s constant. For calculations in the cgs system (centimeter, gram, second), specifically for charges measured in statcoulombs, (k) is set to 1 in the appropriate units, which simplifies the equation to:
[
F = frac{{|q_1 q_2|}}{{r^2}}
]
When considering a medium with a dielectric constant (epsilon_r), the force is reduced compared to the force in a vacuum. The modified Coulomb’s law takes the dielectric constant into account:
[
F = frac{{1}}{{4 pi epsilon_0 epsilon_r}} frac{{|q_1 q_2|}}{{r^2}}
]
In the cgs system, (4 pi
See lessTwo charges 1C and -4C exists in air. What is the direction of force?
The direction of the force between two charges, such as 1C and -4C existing in air, can be determined using Coulomb's law, which states that like charges repel each other and unlike charges attract each other. Since one charge is positive (1C) and the other is negative (-4C), the force between themRead more
The direction of the force between two charges, such as 1C and -4C existing in air, can be determined using Coulomb’s law, which states that like charges repel each other and unlike charges attract each other. Since one charge is positive (1C) and the other is negative (-4C), the force between them would be attractive. Thus, the direction of the force on each charge would be towards the other charge. Specifically, the 1C charge would experience a force in the direction of the -4C charge, and similarly, the -4C charge would experience a force in the direction of the 1C charge.
See less