A conductor with an inner cavity, shown in figure below, carries a total charge of plus 14 mu C. Within the cavity, insulated from the conductor, is a point charge of negative 10 mu C.q13.png
How much charge is on the outer surface of the conductor?
plus 4 mu C
plus 10 mu C
negative 24 mu C
None of the above
plus 14 mu C
The total charge on the conductor includes both the charge within the cavity and the charge on the outer surface.
The charge within the cavity is negative 10 μC.
The total charge on the conductor is plus 14 μC.
Therefore, the charge on the outer surface of the conductor must be the difference between the total charge and the charge within the cavity.
Charge on the outer surface = Total charge - Charge within the cavity
= 14 μC - (-10 μC)
= 14 μC + 10 μC
= 24 μC
So, the charge on the outer surface of the conductor is 24 μC.
The correct answer is negative 24 μC.
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
The magnitude of the electric force between two charges can be calculated using Coulomb's Law:
F = k * |q1| * |q2| / r^2
where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, both charges are electrons (negative 1.6 x 10^-19 C) and the proton is positive (1.6 x 10^-19 C). The distance between them is given as 2 x 10^-10 m.
Plugging in the values:
F = (8.99 x 10^9 N m^2 / C^2) * |(-1.6 x 10^-19 C)| * |(-1.6 x 10^-19 C)| / (2 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2 x 10^-10 m)^2
F ≈ 5.75 x 10^-8 N
Therefore, the magnitude of the net electric force that the two electrons will exert on the proton is approximately 5.75 x 10^-8 N.
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
8.63 cross times 10 to the power of negative 9 space end exponent N
4.98 cross times 10 to the power of negative 9 space end exponent N
9.29 cross times 10 to the power of negative 5 space end exponent N
None of the above
2.87 cross times 10 to the power of negative 9 space end exponent N
To find the magnitude of the net electric force between the two electrons and the proton, we can use Coulomb's Law:
F = k * |q_e| * |q_p| / r^2
where F is the force, k is the electrostatic constant, q_e and q_p are the charges of the electrons and proton respectively, and r is the distance between them.
Plugging in the values:
F = (8.99 x 10^9 N m^2 / C^2) * |(-1.6 x 10^-19 C)| * |(1.6 x 10^-19 C)| / (2 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2 x 10^-10 m)^2
F ≈ 1.837 x 10^-8 N
Therefore, the magnitude of the net electric force that the two electrons will exert on the proton is approximately 1.837 x 10^-8 N.
None of the given options match this value.
The work done to bring a unit positive charge from infinity to a point in an electric field is referred to as:
Electric dipole moment
Electric field intensity
None of the above
The total energy of the point charge
Electric potential
The work done to bring a unit positive charge from infinity to a point in an electric field is referred to as electric potential.
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?
E equals 5 space fraction numerator text v end text over denominator m end fraction
E equals 0.56 space fraction numerator text v end text over denominator m end fraction
None of the above
E equals 25 space fraction numerator text v end text over denominator m end fraction
E equals 0.75 space fraction numerator text v end text over denominator m end fraction
To find the magnitude of the electric field at a point:
E = - ∇V
where E is the electric field, ∇ is the gradient operator, and V is the electric potential.
Taking the partial derivatives of V with respect to x and y:
∂V/∂x = 8v/m^2 * x
∂V/∂y = -18v/m^2 * y
Substituting the given coordinates (1/2, 1/6) into the partial derivatives:
∂V/∂x = 8v/m^2 * (1/2) = 4v/m^2
∂V/∂y = -18v/m^2 * (1/6) = -3v/m^2
Therefore, the electric field at the point (1/2, 1/6) is:
E = - (∂V/∂x, ∂V/∂y) = - (4v/m^2, -3v/m^2) = (-4v/m^2, 3v/m^2)
The magnitude of the electric field is:
|E| = sqrt((-4v/m^2)^2 + (3v/m^2)^2) = sqrt(16v^2/m^4 + 9v^2/m^4) = sqrt(25v^2/m^4) = 5v/m^2
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 5v/m.
The correct answer is E equals 5 fraction v over m.
A flat sheet of paper of area 0.40 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 15 N/C. Find the magnitude of the electric flux through the sheet.
text 3 N.m end text squared divided by C
text 6 N.m end text squared divided by C
text 5.2 N.m end text squared divided by C
None of the above
text 0 N.m end text squared divided by C