Consider a negative charge located between two parallel metal plates that sets up a uniform, downward electric field with magnitude E.
If the charge moves opposite the direction of the electric field,
The electric potential energy increases
The field does a positive work on the charge
The field does a negative work on the charge
None of the above
The electric potential energy decreases
The correct answer is: The electric potential energy decreases.
When a negative charge moves opposite to the direction of the electric field, it is moving from a region of higher electric potential to a region of lower electric potential. Therefore, its electric potential energy decreases.
A charge Q is enclosed by a gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
decrease to one-fourth of its initial value.
The electric flux through a closed surface is given by the formula:
Φ = Q / ε₀
Where Φ is the electric flux, Q is the charge enclosed by the surface, and ε₀ is the permittivity of free space.
The electric flux is proportional to the charge enclosed. If the radius of the Gaussian surface is doubled, the enclosed charge remains the same, so the electric flux will be reduced. Mathematically, the electric flux is inversely proportional to the square of the radius:
Φ ∝ 1 / R²
If the radius is doubled, the electric flux will decrease by a factor of (1/2)² = 1/4, or one-fourth of its initial value.
A point charge q equals plus 6 n Cis surrounded by an imaginary sphere of radius r equals 40 space c mcentered on the charge as shwon in figure below.
q11.png
Calculate the resulting electric flux through the sphere.
To calculate the electric flux through the imaginary sphere surrounding the point charge, we can use Gauss's Law. Gauss's Law states that the electric flux through any closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).
The formula for electric flux through a closed surface is:
Φ = Qenc / ε₀
where Φ is the electric flux, Qenc is the charge enclosed by the surface, and ε₀ is the permittivity of free space.
In this case, the charge enclosed is the charge q itself, which is +6 nC (nanocoulombs). The permittivity of free space, ε₀, is approximately equal to 8.85 x 10^-12 C²/(N·m²).
So, substituting the values into the formula:
Φ = (6 nC) / (8.85 x 10^-12 C²/(N·m²))
Converting nanocoulombs to coulombs (1 nC = 10^-9 C):
Φ = (6 x 10^-9 C) / (8.85 x 10^-12 C²/(N·m²))
Simplifying:
Φ = (6 x 10^-9) / (8.85 x 10^-12 N·m²/C²)
Φ ≈ 6.78 x 10^2 N·m²/C
Therefore, the resulting electric flux through the imaginary sphere is approximately 6.78 x 10^2 N·m²/C.
The figure below shows an arrangement of two electrons each 2 cross times 10 to the power of negative 10 end exponent m from a proton. What is the magnitude of the net electric force they will exert on the proton? q subscript e equals negative 1.6 cross times 10 to the power of negative 19 space end exponent C; q subscript p equals plus 1.6 cross times 10 to the power of negative 19 space end exponent C
q4.png
To calculate the magnitude of the net electric force exerted by the two electrons on the proton, we can use Coulomb's Law. Coulomb's Law states that the magnitude of the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * |q1 * q2| / r²
where F is the electric force, k is the electrostatic constant (approximately 8.99 x 10^9 N·m²/C²), q1 and q2 are the charges, and r is the distance between them.
In this case, the charge of each electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19 C. The distance between them is given as 2 x 10^-10 m.
So, substituting the values into the formula:
F = (8.99 x 10^9 N·m²/C²) * |-1.6 x 10^-19 C * 1.6 x 10^-19 C| / (2 x 10^-10 m)²
Simplifying:
F = (8.99 x 10^9 N·m²/C²) * (2.56 x 10^-38 C²) / (4 x 10^-20 m²)
F = (8.99 x 10^9 N·m²/C²) * (2.56 x 10^-38 C² / 4 x 10^-20 m²)
F = (8.99 x 10^9 N·m²/C²) * (6.4 x 10^-19 N·m²/C²)
F ≈ 5.76 x 10^-9 N
Therefore, the magnitude of the net electric force exerted by the two electrons on the proton is approximately 5.76 x 10^-9 N.
Which of the following is a vector?
Electric potential energy
Electric field
None of the above
Electric potential
Equipotential lines
The electric field is a vector quantity.
The electric potential energy is a scalar quantity.
The electric potential is a scalar quantity.
Equipotential lines are a graphical representation of electric potential and are not vectors themselves.
Therefore, the answer is: Electric field.
Question 13
3
Points
Two parallel plates are charged to produce a potential difference of 60 V. If the separation between the plates is 6 cm as shown in figure below.
Calculate the magnitude of the electric field in the space between the plates.
q18.png
1000 space text V/m end text
10 space text V/m end text
0.1 space text V/m end text
360 space text V/m end text
None of the above
To calculate the magnitude of the electric field in the space between the plates, we can use the formula:
E = V / d
where E is the electric field, V is the potential difference, and d is the separation between the plates.
In this case, the potential difference is given as 60 V and the separation between the plates is 6 cm (which is equal to 0.06 m).
So, substituting the values into the formula:
E = 60 V / 0.06 m
E = 1000 V/m
Therefore, the magnitude of the electric field in the space between the plates is 1000 V/m.
Calculate the magnitude of the electric field at a point P which is 30 cm to the right of a point charge Q equals negative 3 cross times 10 to the power of negative 6 end exponent space text C end text
To calculate the magnitude of the electric field at point P, we can use Coulomb's Law. Coulomb's Law states that the electric field at a point due to a point charge is proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charge and the point.
The formula for the electric field due to a point charge is:
E = k * |q| / r²
where E is the electric field, k is the electrostatic constant (approximately 8.99 x 10^9 N·m²/C²), |q| is the magnitude of the charge, and r is the distance between the charge and the point.
In this case, the charge Q is -3 x 10^-6 C and the distance from the charge to point P is 30 cm (which is equal to 0.3 m).
So, substituting the values into the formula:
E = (8.99 x 10^9 N·m²/C²) * |-3 x 10^-6 C| / (0.3 m)²
Simplifying:
E = (8.99 x 10^9 N·m²/C²) * (3 x 10^-6 C) / (0.3 m)²
E = (8.99 x 10^9 N·m²/C²) * (3 x 10^-6 C / 0.09 m²)
E = (8.99 x 10^9 N·m²/C²) * (3 x 10^-6 / 0.09) N·m²/C
E = (8.99 x 10^9 N·m²/C²) * (3.33 x 10^-5) N·m²/C
E ≈ 299,700 N/C
Therefore, the magnitude of the electric field at point P is approximately 299,700 N/C.
The work done to bring a unit positive charge from infinity to a point in an electric field is referred to as:
The work done to bring a unit positive charge from infinity to a point in an electric field is referred to as the electric potential.
Which one of the following is NOT a property of equipotential surface?
The direction of the equipotential surface is from low potential to high potential.
Two equipotential surfaces can never intersect.
The electric field is always perpendicular to an equipotential surface.
None of the above
No work is done in moving a test charge from one point to another on an equipotential surface.
The statement "The direction of the equipotential surface is from low potential to high potential" is NOT a property of an equipotential surface. The direction of an equipotential surface is always perpendicular to the electric field lines and does not have a specific direction from low to high potential or vice versa.
The electric potential at points in xy plane is given by V left parenthesis x comma y right parenthesis equals open parentheses 4 fraction numerator text v end text over denominator m squared end fraction close parentheses x squared minus open parentheses 9 fraction numerator text v end text over denominator m squared end fraction close parentheses y squared
What is the magnitude of the electric field at the point open parentheses 1 half semicolon 1 over 6 close parentheses?
To find the magnitude of the electric field at a point, we can take the negative gradient of the electric potential function. The electric field is given by:
E = -∇V
where E is the electric field and ∇ is the gradient operator.
First, let's find the partial derivatives of the electric potential function with respect to x and y:
∂V/∂x = (8V/m^2)x
∂V/∂y = (-18V/m^2)y
Next, let's evaluate the partial derivatives at the point (1/2, 1/6):
∂V/∂x = (8V/m^2)(1/2) = 4V/m^2
∂V/∂y = (-18V/m^2)(1/6) = -3V/m^2
Now we have the components of the electric field at the point (1/2, 1/6) as E_x = 4V/m^2 and E_y = -3V/m^2.
Finally, we can find the magnitude of the electric field by taking the square root of the sum of the squares of the components:
|E| = √(E_x^2 + E_y^2)
= √((4V/m^2)^2 + (-3V/m^2)^2)
= √(16V^2/m^4 + 9V^2/m^4)
= √(25V^2/m^4)
= 5V/m^2
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 5V/m^2.