Transcribed Image Text: 5. The Electric Field Due to a Charged Disk The disk has a radius R and a uniform charge density o. Find the electric field E at P. = odA Choose dq as a ring of radius r, dq dE P• (dr r R 1 dqz dE = 4tɛ, (z² +r²)ª 4TE. Integrate to find the electric field E at P below. (Step by step to show your work) Z E = 1- (charged disk) Vz? +R? 280
The problem at hand is to find the electric field due to a charged disk at a point P located below the disk. The disk has a radius R and a uniform charge density o. We can start by considering an infinitesimally small ring on the disk with radius r. Let dq be the charge on this small ring.
dE represents the electric field due to this small ring at point P. To find this, we can start by considering a small element of area on the ring, denoted by dA. The charge on this element of area, which is dq, can be expressed as dq = o*dA, where o is the uniform charge density of the disk.
Now, let’s consider the distance between the point P and the small element of area dq. This distance, denoted by r, can be visualized as the distance from the center of the ring to point P. We can use the Pythagorean theorem to relate r, the distance from the center of the ring to P, to the distance z, which is the perpendicular distance from the disk to P, and the radius R of the disk. The relationship between r, z, and R is given by the equation r² = z² + R².
The electric field dE due to the small element of area dq at point P is given by the equation dE = (k*dq)/(r²), where k is the Coulomb’s constant and r² is the distance between dq and P. Substituting the expression for dq in terms of o and dA, we can rewrite the electric field as dE = (k*o*dA)/(r²).
To find the total electric field E at point P due to the entire disk, we need to integrate the electric field over the entire disk. The integration is performed over r and z, ranging from -∞ to +∞, and from 0 to R, respectively. This is done to consider all possible values of r and z for the small element of area dq on the disk.
The integral expression for the electric field E at point P is:
E = ∫∫ (k*o*dA)/(r²),
where the double integral is taken over the disk.
To evaluate this double integral, we need to express dA and r² in terms of the variables r and z. The infinitesimal area element dA can be written as dA = 2πr*dr, where 2πr is the circumference of the small ring with radius r, and dr is the infinitesimal change in r. Substituting this expression for dA and the equation r² = z² + R² into the integral expression for E, we have:
E = ∫∫ (k*o*2πr*dr)/(z² + R²),
where the double integral is taken over the disk.
Now, we can simplify the integral and evaluate it step by step to find the electric field E at point P. Before performing the integral, notice that the expression (z² + R²) in the denominator does not depend on r. This allows us to factor this expression out of the integral:
E = (k*o*2π)/(z² + R²) ∫∫ r*dr,
where the double integral is taken over the disk.
The remaining integral can be evaluated using the limits of integration. Since we are integrating with respect to r, we need to express the limits of integration for r in terms of the variables z and R. The limits of integration for r depend on the value of z, and the maximum value of r is R due to the finite size of the disk. Therefore, the limits of integration for r are 0 to R.
Using these limits of integration, we can evaluate the remaining integral:
E = (k*o*2π)/(z² + R²) ∫[0, R] r*dr.
E = (k*o*2π)/(z² + R²) * (1/2) * R².
Simplifying this expression further, we can write the electric field E at point P due to the charged disk as:
E = (k*o*π*R²)/(2*(z² + R²)).