An object with charge q_1 3nC is at the origin. An object with charge q_2 = 2nC

Question

An object with charge q_1 3nC is at the origin. An object with charge q_2 = 2nC is at the point 4cmx. Consider the electric field at a point P at (4cm, 2cm, 0). Draw the system and the direction of the electric field from each charge at point P You will have to approximate the length of the field vectors Draw the resultant field vector. Compute the electric field due to charge 1 at point P, E_1p. Compute the electric field due to charge 2 at point P, E_2p. Compute the total electric field at point P.

Explanation / Answer

We use the expression for the electric field

E = k q/r2

Data

. q1 = 3 10-9 C

. q2= 2 10-9 C

. d = 4 cm = 4 10-2 m

We calculate each case separately

.a) p= 4 10-2 m      el punto está sobre la carga 2

E1 = k q1 /p2             E1= 9 109   3 10-9 /(4 10-2)2        E1 = 1.69 104 N/C

E2= k q1/ (p-d)2     if we make direct E 2 calculation gives infinite as the charge and the calculation point match                                                                E2

                                                      …………………………………> may be in the opposite direction

.q1                                            q2 …………..>

      .                                            p=d      E1

                                                          ………………………………………………..>

                                                                Et= E1 +E2

We must emphasize that this configuration is unstable because any movement of the load makes the separate field 2 and can not return to the initial configuration

. b)   p=2 cm = 2 10-2 m

                                 E2             E1

                          < ………… ………….>

. q1                                                                     q2

    0                                P=d/2                             d

                               Et = -E2 +E1

                                           ….>

. r =p

E1= 9 109   3 10-9 / (2 10-2)2         E1 = 6.75 104 N/C     directed to the right

E2 = 9 109   2 10-9/ (2 10-2)2         E2 = 4.5   104 N/C    directed to the left

Et = - E2 + E1 = (-4.5 + 6.75) 104 N/C   Et= 2.25 104 N/C    directed to the right

.c) p= 0 cm      placed over the first charge

E1      direct calculation would infinite as the distance from the zero point shits

E2=   9 109 2 10-9/(4 10-2)2        E2 = 1.125 104 N/C      directed to the left

E2

           <………….    

.                      +q1                                                       +q2

                       0=p                                                       d

   <………………

Et= -E2 +E1

This configuration is also unstable so any movement of the load test p makes the field 2 separate and can not return to the initial configuration.

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