Field due to q1 and q2 is cancelled because of symmetry.
Net electric field E = kq3/d^2 + kq4/[2d]^2
= k*2e/d^2 + k*-8e/4d^2
= 0 Answer
Chapter 22, Problem 008 In the figure the four particles are fixed in place and have...
Chapter 22, Problem 008 In the figure the four particles are fixed in place and have charges qī-92-3e, q3-3e, and q4-12e. D stance d-3.50 ㎛ what is the magnitude of the net electric field at point P due to the particles? 91 9s 2 Number the tolerance is +/-596 Click if you would like to Show Work for this question: Open Show Work Units We were unable to transcribe this image
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r.uni Your answer is partially correct. Try again. In the figure the four particles are fixed in place and have charges q of the net electric field at point P due to the particles? 92-3e, 93-2e, and 4 -Be. Distance d 4.73 um. What is the magnitude on N/C or V/m
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94 91 93 92 In the figure above, the four particles are fixed in place and have charges q_1 = +5e, q_2 = -5e,q_3 = +3e and q_4 = -12e, where e is the magnitude of the charge of electron. Distance N/Cb)What is the electric potential at point P due to the particles? d = 5 micrometers. a)What is the magnitude of the net electric field at point P due to the particles?(Hint: choose a convenient coordinate system.) Volts