Question

Three dipoles are arranged in the x-y plane. The positive and negative charges of dipole 1 is located at (d,d,0) and (-d,-d,0), respectively. For dipole 2, the charges are located at (d,0,0) and (-d,0,0). For dipole 3, they are located at (0,d,0) and (0,-d,0). Assume for all charges that Iq-2 nC and d = 3.0 mm 1. Find an expression for the total dipole moment vector of the system, expressed in rectangular coordinates Find an equation that represents the electric potential due to this net dipole moment for all ที่ (everywhere). Notice that the observation point is expressed as ř in spherical coordinates while the result in (a) is expressed in rectangular coordinates. a. b. c. How much work is done to move a test charge qtest-4.0 nC from To(08 + 1y + 52) meters to = (42 + 0ỹ + 02) meters ? [hint: calculate the electrical potential (voltage) difference between the two points and then multiply this voltage difference by the test charge value to obtain the work. Your answer should be a numerical value in units of Joules.]

Answer 1

Here 1st of all calculate dipole moment for single dipole in standard form P vectod. Which is multiple of charge and displacement between chargeses. Direction from -ve to +ve. Now we calculate vector sum of p1,p2 and p3 which gives p resultant.

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now we calculate electric potential V by single dipole in standard form. Then convert in vector form . Which is finally dot product of dipole moment and position vector of general point.

"Here 1st of all calculate dipole moment for single dipole in standard form P vector. Which is multiple of charge and displacement between charges. Direction from -ve to + ve. Now we calculate vector sum of p1, p2 and p3 which gives p resultant. Ans: we know that for single dipole rightarrow (a) dimple moment given by [P^rightarrow = q d^rightarrow] Here d^rightarrow is from (-q) to (+ q). For three dipoles: we take vector sum of individual. i.e P^rightarrow_R = q(2di^+ 2dj^) + q(2di^) + (2dj^)q [P^rightarrow_R = 4qd (i^+ j^)]

for work done in moving a test charge we calculate potential difference between initial to final position.

Then multiply with value of test charge. Which gives desired work done. hence net work done is -47.48 joules.