Electric Fields Calculate the net electric force (magnitude and direction) on a

Question

Electric Fields

Calculate the net electric force (magnitude and direction) on a positive test charge placed at point A in the diagram of charges shown on the right. Each charge, including the test charge, has a magnitude of 1 nC and each small division on the diagram is 0.2 m. Coulomb's law states that the electric force exerted by a point charge qi on a second point charge q2 is F rightarrow 12 = kB 1q2/r2 r12 where ke is Coulomb's constant (KB = 8.99 times 109 Nm2/C2), r is the distance between the charges and r12 is a unit vector directed from q1 to q2. Calculate the electric potential (mgnitude only as it is a scalar quantity ) at point A in the figure above. Identify the types of charges by placing + or - signs in the empty circles for the diagram shown below. Sketch the electric field lines for the diagram. Where is the electric field stronger: at the triangle or the square marker? Why? The figure below shows an electric field extending over three regions, labeled I, II, and III. Are there any isolated charges in the diagram? If no, skip to b. If so, indicate in what region (I, II, or III) they are located. If so, indicate the sign of these changes. In what region (I, II, or III) is the field the strongest? In what region (I, II, or III) is the field the weakest? In what region (I, II, or III) is the field the most uniform? Calculate the electric potential 0.50 m from a 4.5 times 10-4 C point charge.

Explanation / Answer

1.

Net Force = F = 2.2475 * 10^-9 N (Towards Positive charge)

2.

Electric potential at A = kq1/r1 + kq2/r2 + kq3/r3 = 13.485 V

3.

Higher the density of lines, more the electric field. so, hence the electric field is stronger at the triangle.

4.

a.

At position II,

they are negative charges as the electric firld lines are converging at them.

b.

Higher the density of lines, more the electric field. so, hence the electric field is stronger at II.

c.

Region III

d.

Region I

5.

Potential V = kq/r = 8.091 * 10^ 6 Volts

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