+9.70 nC of charge is uniformly distributed along the top half of a thin rod of total length L = 8.00 cm, while -9.70 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 = 21.8 cm from the centre of the rod?
Question
+9.70 nC of charge is uniformly distributed along the top half of a thin rod of total length L = 8.00 cm, while -9.70 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 = 21.8 cm from the centre of the rod?
Solution
The electric field due to a uniformly charged rod at a point along the axis of the rod is given by the formula:
E = kλ/r
where:
- E is the electric field,
- k is Coulomb's constant (9 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 point to the rod.
First, we need to calculate the linear charge density. The total charge on the rod is 0 (since +9.70 nC and -9.70 nC cancel each other out), but the charge is distributed differently on the top and bottom halves. So, we calculate the linear charge density for each half:
λ_top = Q_top / L_top = +9.70 nC / 4.00 cm = +2.425 nC/cm λ_bottom = Q_bottom / L_bottom = -9.70 nC / 4.00 cm = -2.425 nC/cm
Next, we calculate the electric field due to each half at the point. The total electric field will be the vector sum of these two fields. Since the fields due to the top and bottom halves are in opposite directions, we subtract the magnitudes:
E_top = kλ_top / r = (9 x 10^9 N m^2/C^2) * (2.425 x 10^-9 C/m) / (21.8 cm) = 1.00 N/C E_bottom = kλ_bottom / r = (9 x 10^9 N m^2/C^2) * (-2.425 x 10^-9 C/m) / (21.8 cm) = -1.00 N/C
E_total = E_top - E_bottom = 1.00 N/C - (-1.00 N/C) = 2.00 N/C
So, the magnitude of the electric field at the dot is 2.00 N/C.
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