Question

1. (15 points) Three charges are shown below. Find the following values a. What is the electric field at the point labelled A? b. What is the electric potential at the point labelled A? c. What is the electric force on the +10°C charge? -buc A +64 с. 0.3 m 10.3m 0.4 + Oud

Answer 1

The electric field due to a point charge can be determined by using Coulomb's law

Electric field due to positive charge is radially outwardand (which is positive) due to negative charge is radially inward to it(which is negative). Here we have two positive charges and one negative charge. Electric field at point A will be algebraic sum of electric field due to individual charges.

k Q source 9 KQ source E شه qY2 72 where, k 9x109 Nm²/C2 Coulomb's constant

(since there is a symmetry in first two charges, that is - 6microC and +6microC charge, both are at equal distance from A and both have charge of same magnitude. Therefore electric field at A due to these two charges are equal in magnitude and opposite in direction, therefore cancels each other.Hence only electric field at A due to +10microC charge need to calculate.This is possible because electric field is a vector quantity .)

- Guc F4 + 6HC 0.3m 0.3m E3' 10.4m O +10C Electric field at A due to charge - 6HC (-6)x106 (0.3)2 E, = -6x10°NIC E = ko u 9x109 x Electric field at A due to charge +GHC Ea = 9x109x GX106 (0.3)2 EQ = 6x105N/C Electric field at A due to charge +10 HC E3 10x106 9x1097 (0.412 E3 = 56.25x10* NIC Therefore, Electric field at point A El+Ea +E3 -6x10 + 6x109+ 56.25x104 56.25*10* NIC

Electric potential is a scalar quantity. The electric potential at a point in an electric field is defined as the amount of work done in moving a unit positive charge from infinity to that point along any path when the electrostatic forces are applied. Suppose that a positive charge is placed at a point. The charge placed at that point will exert a force due to the presence of an electric field. The electric potential at any point at a distance r from a charge q is

The electric force on the charge 10microC is the vector sum of electric force caused by the charges 6microC and - 6microC. ​​​

V= دامن ATE, r Electric potential due to multiple charges Qa + Q3] 1 + VE 47Eo ri X2 Therefore, electric potential at A, V Gxlo Gx17c_ 10_x10 9x109 0.94 22 xlo] 0.3 0.3 4x10² 10 22 12 10 + 9 0.4 ] 9x109x106 [ 0.3 9x103 [123 58 5X103 J/C (12x0.4) + 3 0.la

According to Coulomb's law, the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.The force is along the straight line joining them. If the two charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.

FB - GMC A 0.3 m А 0.3m +6HC B Bi Bat 0.4m 0.5 m 0.5m 10.3+0.42 = 0.5m AC = BC- +10°C.. Өa FB 1 1 IFBI 9,92 I FAI (6x106) (10x106) (0.5) 40E. 72 4TE. (6x106) (10x106) 4TEO (0,5) 2.16N 12 9x109 x 6 x 60 x 10 25xlöa 2.16N Resultant force on charge + 10 MC is lēlu JFP+ FALIP 2FgFa cosa Since both have same magnitude IRI 2.16 / 2+ acoso to find o , o = 0.02

0.4 from figure, sino = - tes 0.5 5 0,= sin(4) = 53.130 Ba = sin' () 53.130 0 - Oitoa = 2x 53.130 106.26° IRI 2.16 Ja+acos 106260 IRI = 2.592 N
Therefore force on charge 10microC is 2.592N