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+4 o +1 o -3 o The E-Field formed by multiple charged parallel plates is the vector sum of all three fields. Let us begin by looking at the E-field formed at point A. The E-field due to the first plate has a magnitude of 40/2ɛ, or 20/ɛo. The first plate has a positive charge. So the E-field at point A is away from the plate to the right. The direction is to the right, so we will say the E-field due to the first plate is: +2 o/ɛ, . Computer's answer now shown above. Incorrect. Tries 2/2 Previous Tries The E-field due to the second plate has a magnitude of 10/2ɛ, or 1/20/Eo. The second plate has a positive charge. So the E-field at point A is away from the plate - to the left. The direction is to the left, so we will say the E-field due to the second plate is: -1/2 ơ/ɛo . Computer's answer now shown above. You are correct. Your receipt no. is 159-419 ? Previous Tries The E-field due to the third plate has a magnitude of 3o/2ɛ, or 3/20/ɛo. The third plate has a negative charge. So the E-field at point A is towards the plate - to the right. The direction is to the right, so we will say the E-field due to the second plate is: +3/2 o/ɛ, . Computer's answer now shown above. You are correct. Your receipt no. is 159-3638 ? Previous Tries Adding together the E-field of all three plates will now give us the total E-field at point A: Submit Answer Tries 0/2

+4 o +1 o -3 o The E-Field formed by multiple charged parallel plates is the vector sum of all three fields. Let us begin by looking at the E-field formed at point A. The E-field due to the first plate has a magnitude of 40/2ɛ, or 20/ɛo. The first plate has a positive charge. So the E-field at point A is away from the plate to the right. The direction is to the right, so we will say the E-field due to the first plate is: +2 o/ɛ, . Computer's answer now shown above. Incorrect. Tries 2/2 Previous Tries The E-field due to the second plate has a magnitude of 10/2ɛ, or 1/20/Eo. The second plate has a positive charge. So the E-field at point A is away from the plate - to the left. The direction is to the left, so we will say the E-field due to the second plate is: -1/2 ơ/ɛo . Computer's answer now shown above. You are correct. Your receipt no. is 159-419 ? Previous Tries The E-field due to the third plate has a magnitude of 3o/2ɛ, or 3/20/ɛo. The third plate has a negative charge. So the E-field at point A is towards the plate - to the right. The direction is to the right, so we will say the E-field due to the second plate is: +3/2 o/ɛ, . Computer's answer now shown above. You are correct. Your receipt no. is 159-3638 ? Previous Tries Adding together the E-field of all three plates will now give us the total E-field at point A: Submit Answer Tries 0/2

Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter23: Electric Fields
Section: Chapter Questions
Problem 23.11OQ: Three charged particles are arranged on corners of a square as shown in Figure OQ23.11, with charge...
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+4 o +1 o -3 o A The E-Field formed by multiple charged parallel plates is the vector sum of all three fields. Let us begin by looking at the E-field formed at point A. The E-field due to the first plate has a magnitude of 40/2ɛ, or 20/E0. The first plate has a positive charge. So the E-field at point A is away from the plate to the right. The direction is to the right, so we will say the E-field due to the first plate is: +2 o/ɛ, . Computer's answer now shown above. Incorrect. Tries 2/2 Previous Tries The E-field due to the second plate has a magnitude of l0/2ɛ, or 1/20/E0. The second plate has a positive charge. So the E-field at point A is away from the plate to the left, The direction is to the left, so we will say the E-field due to the second plate is: -1/2 o/ɛo . Computer's answer now shown above. You are correct. Your receipt no. is 159-419 ? Previous Tries The E-field due to the third plate has a magnitude of 30/2ɛ, or 3/20/ɛo. The third plate has a negative charge. So the E-field at point A is towards the plate to the right. The direction is to the right, so we will say the E-field due to the second plate is: +3/2 0/ɛo . Computer's answer now shown above. You are correct. Your receipt no. is 159-3638 ? Previous Tries Adding together the E-field of all three plates will now give us the total E-field at point A: Submit Answer Tries 0/2 Notice that the E-fields at point B will have the same magnitude but the direction and therefore the sign may change. Use the same method to now calculate the total E-field at point B:
Transcribed Image Text:+4 o +1 o -3 o A The E-Field formed by multiple charged parallel plates is the vector sum of all three fields. Let us begin by looking at the E-field formed at point A. The E-field due to the first plate has a magnitude of 40/2ɛ, or 20/E0. The first plate has a positive charge. So the E-field at point A is away from the plate to the right. The direction is to the right, so we will say the E-field due to the first plate is: +2 o/ɛ, . Computer's answer now shown above. Incorrect. Tries 2/2 Previous Tries The E-field due to the second plate has a magnitude of l0/2ɛ, or 1/20/E0. The second plate has a positive charge. So the E-field at point A is away from the plate to the left, The direction is to the left, so we will say the E-field due to the second plate is: -1/2 o/ɛo . Computer's answer now shown above. You are correct. Your receipt no. is 159-419 ? Previous Tries The E-field due to the third plate has a magnitude of 30/2ɛ, or 3/20/ɛo. The third plate has a negative charge. So the E-field at point A is towards the plate to the right. The direction is to the right, so we will say the E-field due to the second plate is: +3/2 0/ɛo . Computer's answer now shown above. You are correct. Your receipt no. is 159-3638 ? Previous Tries Adding together the E-field of all three plates will now give us the total E-field at point A: Submit Answer Tries 0/2 Notice that the E-fields at point B will have the same magnitude but the direction and therefore the sign may change. Use the same method to now calculate the total E-field at point B:
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