Two charged disks that are 35.2 cm apart face each other. The radius of each disk is 16.0 cm, the charge on the left disk is -2.10 nC, and the charge on the right disk is +2.10 nC. What is the magnitude of the electric field at the midpoint between the two disks?
Question
Two charged disks that are 35.2 cm apart face each other. The radius of each disk is 16.0 cm, the charge on the left disk is -2.10 nC, and the charge on the right disk is +2.10 nC. What is the magnitude of the electric field at the midpoint between the two disks?
Solution
To solve this problem, we need to use the formula for the electric field due to a point charge, which is given by:
E = kQ/r^2
where E is the electric field, k is Coulomb's constant (9.0 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
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First, convert the charge from nC to C. 1 nC = 10^-9 C. So, Q1 = Q2 = 2.10 nC = 2.10 x 10^-9 C.
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The distance from each disk to the midpoint is half of the total distance between the disks, so r = 35.2 cm / 2 = 17.6 cm. Convert this to meters: r = 17.6 cm = 0.176 m.
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Now, calculate the electric field due to each disk at the midpoint. Since the charges are equal in magnitude but opposite in sign, their electric fields at the midpoint will also be equal in magnitude but opposite in direction. Therefore, we can calculate the electric field due to one disk and then double it to get the total electric field.
E1 = kQ1/r^2 = (9.0 x 10^9 N m^2/C^2) * (2.10 x 10^-9 C) / (0.176 m)^2 = 607 N/C
- The total electric field at the midpoint is then 2E1 = 2 * 607 N/C = 1214 N/C.
So, the magnitude of the electric field at the midpoint between the two disks is 1214 N/C.
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