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Electric field intensity due to infinite sheet of charge σ is
The electric field intensity ( E ) due to an infinite sheet of charge with surface charge density ( sigma ) is given by the formula:[E = frac{sigma}{2epsilon_0}]where ( epsilon_0 ) is the permittivity of free space. The field is uniform and points away from the sheet if the charge is positive, or toRead more
The electric field intensity ( E ) due to an infinite sheet of charge with surface charge density ( sigma ) is given by the formula:
[
E = frac{sigma}{2epsilon_0}
]
where ( epsilon_0 ) is the permittivity of free space. The field is uniform and points away from the sheet if the charge is positive, or toward the sheet if the charge is negative.
See lessElectric field of an infinitely long conductor of charge density λ, is given by E = λ/(2πεh).aN. State True/False.
False.
False.
See lessThe field intensity of a charge defines the impact of the charge on a test charge placed at a distance. It is maximum at d = 0cm and minimises as d increases. State True/False
False
False
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 (E) of 40 V/cm at a distance (r) of 30 cm in a vacuum, we can use the formula for the electric field due to a point charge:[E = frac{k cdot |q|}{r^2}]Where:- (E) is the electric field strength (in V/m),- (k) is Coulomb's constant, approximaRead more
To find the charge that produces an electric field strength (E) of 40 V/cm at a distance (r) of 30 cm in a vacuum, we can use the formula for the electric field due to a point charge:
[
E = frac{k cdot |q|}{r^2}
]
Where:
– (E) is the electric field strength (in V/m),
– (k) is Coulomb’s constant, approximately (8.99 times 10^9 , text{N m}^2/text{C}^2),
– (q) is the charge (in coulombs),
– (r) is the distance from the charge (in meters).
Step 1: Convert units
– (E = 40 , text{V/cm} = 4000 , text{V/m})
– (r = 30 , text{cm} = 0.3 , text{m})
Step 2: Rearrange the formula to solve for charge (q)
[
|q| = frac{E cdot r^2}{k}
]
Step 3: Substitute the values
[
|q| = frac{4000 , text{V/m} cdot (0.3 , text{m})^2}{8.99
See lessWhat is the electric field intensity at a distance of 20cm from a charge 2 X 10-6 C in vacuum?
To calculate the electric field intensity (E) at a distance (r) from a point charge (Q) in vacuum, we use the formula:[ E = frac{k cdot |Q|}{r^2} ]where:- ( E ) is the electric field intensity,- ( k ) is Coulomb's constant, approximately ( 8.99 times 10^9 , text{N m}^2/text{C}^2 ),- ( |Q| ) is the mRead more
To calculate the electric field intensity (E) at a distance (r) from a point charge (Q) in vacuum, we use the formula:
[ E = frac{k cdot |Q|}{r^2} ]
where:
– ( E ) is the electric field intensity,
– ( k ) is Coulomb’s constant, approximately ( 8.99 times 10^9 , text{N m}^2/text{C}^2 ),
– ( |Q| ) is the magnitude of the charge, and
– ( r ) is the distance from the charge.
Given:
– ( Q = 2 times 10^{-6} ) C,
– ( r = 20 ) cm = ( 0.2 ) m.
Now, substituting the values into the formula:
[ E = frac{8.99 times 10^9 , text{N m}^2/text{C}^2 cdot 2 times 10^{-6} , text{C}}{(0.2 , text{m})^2} ]
Calculating the denominator:
[(0.2)^2 = 0.04 , text{m}^2]
Now substituting back:
[ E = frac{8.99 times 10^9 cdot
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 the two charges, we can use Coulomb's Law, which is given by the formula:[F = k frac{|q_1 q_2|}{r^2}]where:- ( F ) is the force between the charges (0.1 N),- ( k ) is Coulomb's constant (( 8.99 times 10^9 , text{Nm}^2/text{C}^2 )),- ( q_1 ) and ( q_2 ) are the magniRead more
To determine the distance between the two charges, we can use Coulomb’s Law, which is given by the formula:
[
F = k frac{|q_1 q_2|}{r^2}
]
where:
– ( F ) is the force between the charges (0.1 N),
– ( k ) is Coulomb’s constant (( 8.99 times 10^9 , text{Nm}^2/text{C}^2 )),
– ( q_1 ) and ( q_2 ) are the magnitudes of the charges (( 2 times 10^{-7} , C ) and ( 4.5 times 10^{-7} , C ), respectively),
– ( r ) is the distance between the charges.
Rearranging the formula to solve for ( r ) gives:
[
r = sqrt{k frac{|q_1 q_2|}{F}}
]
Now we can substitute the given values into the formula:
[
r = sqrt{8.99 times 10^9 , frac{(2 times 10^{-7})(4.5 times 10^{-7})}{0.1}}
]
Calculating the numerator:
[
|q_1 q_2| = (2 times 10^{-7})(4.5 times 10
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 negatively charged dielectric balls, we can use Coulomb's law, which describes the electrostatic force between two point charges. 1. Given:- Charge of each ball, ( q = 1 , mu C = 1 times 10^{-6} , C )- Mass of each ball, ( m = 10 , g = 0.01 , kg )- The equationRead more
To find the separation between the two negatively charged dielectric balls, we can use Coulomb’s law, which describes the electrostatic force between two point charges.
1. Given:
– Charge of each ball, ( q = 1 , mu C = 1 times 10^{-6} , C )
– Mass of each ball, ( m = 10 , g = 0.01 , kg )
– The equation for the electrostatic force between two charges is given by:
[
F = k frac{|q_1 q_2|}{r^2}
]
where ( F ) is the electrostatic force, ( k ) is Coulomb’s constant (approximately ( 8.99 times 10^9 , N m^2/C^2 )), ( r ) is the separation between the charges, and ( q_1 ) and ( q_2 ) are the charges of the two balls.
2. For this configuration:
– Since one ball is restrained, the force acting on the upper ball must balance the gravitational force acting on it.
3. Calculating the gravitational force ( F_g ) acting on the upper ball:
[
F_g = m g
]
where ( g ) (acceleration due to gravity) is approximately ( 9.81 ,
See lessThe Coulomb law is an implication of which law?
The Coulomb law is an implication of Gauss's law.
The Coulomb law is an implication of Gauss’s law.
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 when they are separated by a distance, we can use Coulomb's Law, which is given by the formula:[F = k frac{|q_1 q_2|}{r^2}]where:- ( F ) is the force between the charges,- ( k ) is Coulomb's constant (( 8.99 times 10^9 , text{N m}^2/text{C}^2 )),- ( q_1 ) and (Read more
To find the force between two charges when they are separated by a distance, we can use Coulomb’s Law, which is given by the formula:
[
F = k frac{|q_1 q_2|}{r^2}
]
where:
– ( F ) is the force between the charges,
– ( k ) is Coulomb’s constant (( 8.99 times 10^9 , text{N m}^2/text{C}^2 )),
– ( q_1 ) and ( q_2 ) are the charges,
– ( r ) is the distance between the charges.
Given:
– ( q_1 = 2 , text{nC} = 2 times 10^{-9} , text{C} )
– ( q_2 = -1 , text{nC} = -1 times 10^{-9} , text{C} )
– ( r = 4 , text{cm} = 0.04 , text{m} )
Substituting these values into the formula:
[
F = (8.99 times 10^9) frac{|(2 times 10^{-9})(-1 times 10^{-9})|}{(0.04)^2}
]
Calculating the numerator:
[
|(
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 find the force of interaction between two point charges in a medium, we can use Coulomb's Law in the form:[F = frac{1}{4 pi epsilon} cdot frac{|q_1 q_2|}{r^2}]where:- ( F ) is the force between the charges,- ( epsilon ) is the permittivity of the medium,- ( q_1 ) and ( q_2 ) are the magnitudes ofRead more
To find the force of interaction between two point charges in a medium, we can use Coulomb’s Law in the form:
[
F = frac{1}{4 pi epsilon} cdot frac{|q_1 q_2|}{r^2}
]
where:
– ( F ) is the force between the charges,
– ( epsilon ) is the permittivity of the medium,
– ( q_1 ) and ( q_2 ) are the magnitudes of the charges,
– ( r ) is the distance between the charges.
In this case, stat coulombs need to be converted to SI units for practical calculation. The conversion is:
1 stat coulomb = ( 3.33564 times 10^{-10} ) C.
So:
– ( q_1 = 60 text{ stat coulombs} = 60 times 3.33564 times 10^{-10} ) C,
– ( q_2 = 37.5 text{ stat coulombs} = 37.5 times 3.33564 times 10^{-10} ) C.
The distance is given as 7.5 cm, which is:
[
r = 7.5 text{ cm} = 0.075 text{ m}
]
Next, to find the permittivity ( epsilon ) for the
See less