# Date:8.10.13'm.3mthe origin.(-7.10 Name:Date:8.• Zm10.13'mUse the charges shown:91 = 5 µC92 = -7µC%3D.3m1. Find the value of the electric potential at the origin.(5.10t3 (-2.10yKaiVDart.= r8,99103|4kaz=(6,99:100,2030.7'27=14983= 15000 V0.215000 V.2 Find the electric field at the origin. Draw the electric field vector at the origin on the figure.(6.99.10>2)=1,12.10°(0.2))-1.17.1560.3)"4. 7806.,12X10(8.99.10")(==69927orlt od1.32XW3. What would be the force on q3 = -3 uC if it were placed at the origin?%3D-2.1i +336 53.96 so4. What would bc the potential energy of q3 = -3 µC at the origin?%3D

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Can you help me find the electric field at the origin. I can draw it once I found it.

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Step 1

(a)

Consider k will be the Columbo’s constant, q is the charge and r is the distance.

The formula for the electric potential is:

Step 2

Since the electric potential is the scalar quantity. Therefore, the electric potential at origin will be:

Step 3

(b)

The formula for the electric field is:

Since the electric field is a vector quantity. Therefore, the electric field at origin will be:

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