+8.30 nC of charge is uniformly distributed along the top half of a thin rod of total length L = 4.00 cm, while -8.30 nC of charge is uniformly distributed along the bottom half of the rod, as shown in the figure. What is the magnitude of the electric field at the dot, a distance r = 11.7 cm from the centre of the rod?
Question
+8.30 nC of charge is uniformly distributed along the top half of a thin rod of total length L = 4.00 cm, while -8.30 nC of charge is uniformly distributed along the bottom half of the rod, as shown in the figure. What is the magnitude of the electric field at the dot, a distance r = 11.7 cm from the centre of the rod?
Solution
To solve this problem, we need to use the formula for the electric field due to a line of charge, which is given by:
E = kλ/r
where:
- E is the electric field,
- k is Coulomb's constant (9.0 x 10^9 N m^2/C^2),
- λ is the linear charge density (charge per unit length), and
- r is the distance from the line of charge.
First, we need to calculate the linear charge density for the top and bottom halves of the rod. The total charge on each half is ±8.30 nC, and the length of each half is L/2 = 4.00 cm / 2 = 2.00 cm = 0.02 m. Therefore, the linear charge density is:
λ = Q/L = ±8.30 nC / 0.02 m = ±415 nC/m = ±415 x 10^-9 C/m
Next, we calculate the electric field due to each half of the rod at the point of interest. The distance from the center of the rod to the point is given as r = 11.7 cm = 0.117 m. Therefore, the electric field due to each half is:
E = kλ/r = 9.0 x 10^9 N m^2/C^2 * ±415 x 10^-9 C/m / 0.117 m = ±318.8 N/C
Finally, because the charges on the top and bottom halves of the rod are equal and opposite, their electric fields at the point of interest will also be equal and opposite. Therefore, they will cancel out, and the total electric field at the point will be zero.
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