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.
337.13 space N m squared divided by C
2.4 space cross times 10 to the power of negative 9 end exponent space N m squared divided by C
1.2 space cross times 10 to the power of negative 9 end exponent space N m squared divided by C
677.97 space N m squared divided by C
None of the above
The electric flux through a closed surface is given by the equation:
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.
In this case, the charge enclosed by the sphere is q = 6 nC = 6 * 10^-9 C. The permittivity of free space is ε₀ = 8.85 * 10^-12 N m^2/C^2.
Plugging these values into the equation, we get:
Φ = (6 * 10^-9 C)/(8.85 * 10^-12 N m^2/C^2)
= 677.97 N m^2/C
Therefore, the resulting electric flux through the sphere is 677.97 N m^2/C.
So the correct answer is:
677.97 N m^2/C
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
360 space text V/m end text
10 space text V/m end text
None of the above
1000 space text V/m end text
0.1 space text V/m end text
The magnitude of the electric field between two parallel plates with potential difference V and separation distance d is given by:
E = V/d
In this case, the potential difference is V = 60 V and the separation distance is d = 6 cm = 0.06 m.
Plugging these values into the equation, we get:
E = 60 V / 0.06 m
= 1000 V/m
Therefore, the magnitude of the electric field in the space between the plates is 1000 V/m.
So the correct answer is:
1000 V/m
Three point charges are arranged along the x-axis. Charge q subscript 1 equals plus 6 mu C is at the origin, charge q subscript 2 equals negative 10 mu C is at x equals 0.4 m and charge q subscript 3 equals negative 16 mu C. Where is q subscript 3 located if the net force on q subscript 1 is 14 N in the minusx-direction (negative)?
To determine the location of q3, we can use Coulomb's Law to calculate the net force on q1 caused by q2 and q3 and then use the principle of superposition to find the net force.
Coulomb's Law states that the force between two charges is given by:
F = k * (|q1| * |q2|) / r^2
where F is the force, k is the electrostatic constant (9 × 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
Let's calculate the forces between q1 and q2, and between q1 and q3.
Force between q1 and q2:
F12 = k * (|q1| * |q2|) / r12^2
= 9 × 10^9 N m^2/C^2 * (6 μC * 10 μC) / (0.4 m)^2
Force between q1 and q3:
F13 = k * (|q1| * |q3|) / r13^2
= 9 × 10^9 N m^2/C^2 * (6 μC * 16 μC) / (r13)^2
Since the net force on q1 is given as 14 N in the -x direction, we know that the total force (Fnet) on q1 is the sum of the forces F12 and F13:
Fnet = F12 + F13
Substituting the calculated values, we have:
14 N = (9 × 10^9 N m^2/C^2 * (6 μC * 10 μC) / (0.4 m)^2) + (9 × 10^9 N m^2/C^2 * (6 μC * 16 μC) / (r13)^2)
Let's solve this equation for r13:
14 N = ((54 * 10^-6 C^2) / (0.4 m)^2) + ((86 * 10^-6 C^2) / r13^2)
Multiplying through by r13^2:
14 N * r13^2 = (54 * 10^-6 C^2) + (86 * 10^-6 C^2)
Dividing through by 10^-6 C^2:
14 * 10^6 N * r13^2 = 54 + 86
Let's simplify:
14 * 10^6 N * r13^2 = 140
r13^2 = 140 / (14 * 10^6 N)
r13^2 = 10 / (10^6 N)
r13^2 = 10^-5 / N
Taking the square root of both sides:
r13 = √(10^-5 / N)
Therefore, the location of q3 is at a distance of √(10^-5 / N) from q1 along the x-axis.
To calculate the resulting electric flux through the sphere, we can use Gauss's law. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ε0).
The formula for electric flux is:
Φ = qenclosed / ε0
In this case, the charge enclosed by the sphere is the same as the charge at the center of the sphere. So,
qenclosed = q = 6 nC
The permittivity of free space (ε0) is approximately equal to 8.85 x 10^-12 N m^2/C^2.
Plugging in these values into the formula, we get:
Φ = (6 nC) / (8.85 x 10^-12 N m^2/C^2)
Calculating this expression, we find:
Φ ≈ 6.7797 x 10^9 N m^2/C
Therefore, the resulting electric flux through the sphere is approximately 6.7797 x 10^9 N m^2/C.
None of the given answer choices match this result, so the correct answer is None of the above.
To calculate the resulting electric flux through the sphere, you can use Gauss's Law. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).
In this case, we have a point charge q surrounded by an imaginary sphere. The electric flux through the sphere can be calculated as:
Electric Flux = q / ε₀
First, we need to determine the charge enclosed by the sphere. Since the sphere is centered on the charge, all of its flux lines pass through the sphere. Therefore, the charge enclosed is equal to the given charge q.
Next, we need to determine the permittivity of free space (ε₀), which is a constant value equal to approximately 8.85 x 10^(-12) N m²/C².
Now we can substitute the values into the electric flux equation:
Electric Flux = (6 nC) / (8.85 x 10^(-12) N m²/C²)
Converting the charge from nanocoulombs to coulombs:
Electric Flux = (6 x 10^(-9) C) / (8.85 x 10^(-12) N m²/C²)
Simplifying the expression:
Electric Flux = 677.97 N m²/C
Therefore, the correct answer is 677.97 N m²/C.