Charge +Q = +7.10 nC is uniformly distributed along the right half of a thin rod bent into a semicirle of radius R = 3.60 cm, while charge −Q = −7.10 nC is uniformly distributed along the left half of the rod. What is the magnitude and direction of the electric field at point P, the centre of the circle?
http://www.phys.uri.edu/gerhard/PHY204/tsl329.pdf
Then double that.
I don't get it
To find the electric field at point P, we can divide the problem into two parts: the electric field due to the positive charge and the electric field due to the negative charge.
Let's start with the electric field due to the positive charge. We know that the electric field due to a point charge is given by the equation:
E_positive = k * (Q_positive / r^2)
where E_positive is the electric field, k is the electrostatic constant, Q_positive is the charge, and r is the distance between the charge and the point where we want to find the electric field.
Since the positive charge is distributed along the right half of the semicircle, we can consider a small element on the semicircle (dθ) and find the electric field due to this small element at point P. We can then integrate these electric fields over the entire semicircle to find the total electric field.
The electric field due to the positive charge element can be written as:
dE_positive = k * (dQ_positive / r^2)
where dQ_positive is the charge of the small element.
To find dQ_positive, we divide the total charge (+Q = +7.10 nC) by the total circumference of the semicircle (2πR/2). Thus, dQ_positive = (+Q * dθ) / (2πR/2) = (+Q * dθ) / πR.
Substituting this in the expression for dE_positive, we get:
dE_positive = k * [(+Q * dθ) / πR] / r^2
Now, we can integrate dE_positive over the entire semicircle, from 0 to π radians, to find the total electric field due to the positive charge:
E_positive = ∫[0,π] dE_positive = ∫[0,π] k * [(+Q * dθ) / πR] / r^2 = (k * Q / πR) * ∫[0,π] dθ / r^2
Since r is the radius of the circle and is equal to R (given to be 3.60 cm), we can simplify further:
E_positive = (k * Q / πR) * ∫[0,π] dθ / R^2 = (k * Q / πR) * (θ / R^2) ∣ [0,π] = (k * Q / πR) * (π / R^2) = (k * Q) / R^3
Similarly, we can find the electric field due to the negative charge using the same logic:
E_negative = (k * Q) / R^3
Now, since the electric field is a vector quantity, we need to consider the directions of the two electric fields.
The positive charge is distributed along the right half of the semicircle. At point P, the electric field due to this positive charge will be directed towards the right of the circle.
The negative charge is distributed along the left half of the semicircle. At point P, the electric field due to this negative charge will be directed towards the left of the circle.
Since the magnitudes of the electric fields due to the positive and negative charges are the same, their vector sum will cancel each other out at point P, resulting in a net electric field of zero magnitude.
Therefore, the magnitude of the electric field at point P is zero and its direction is undefined.