Three charged particles are at the corners of an equilateral triangle as shown i

Question

Three charged particles are at the corners of an equilateral triangle as shown in the figure below. (Let q = 2.00 pC, and L = 0.550 m.) 7.00 pC 60.0 4.00 (a) Calculate the electric field at the position of charge q due to the 7.00-uC and -4.00-HC charges Once you calculate the magnitude of the field contribution from each charge you need to add these as vectors. kN/Ci 180 (b) Use your answer to part (a) to determine the force on charge q. If you know the electric field at a particular point, how do you find the force that acts on a charge at that point? mN j Need Help? meen-Lene. 6 Three point charges are located on a circular arc as shown in the figure below. (Taker 3.96 cm. Let to the right be the +x direction and up along the screen be the +y direction.) +3,00 nC -2.00 nC 50 +3.00 nC (a) What is the total electric field at P, the center of the arc? (b) Find the electric force that would be exerted on a -5.02-nC point charge placed atP 9 A charged cork ball of mass 2.70 g is suspended on a light string in the presence of a uniform electric field as shown in the figure below. When E = (3.801 + 6.00 X 105 N/C, the ball is in equilibrium at = 37.0° (a) Find the charge on the ball. b) Find the tension in the string. 10 A horizontal insulating od of length 95-cm and charge 3 is i a plane with a long straight vertical unifo n lne charge. The linear charge density of the long line charge is 2.0 × 10-7 closest to the line charge is 2.0 cm away? Hint: the electric field from an infinitely long straight uniform line charge is )- 00228x N m. What is the electric force on the rod if the e

Explanation / Answer

5 )

a )

q = 2 uC = 2 X 10-6 C

L = 0.55 m

E1 = k q1 i / L2

= 9 X 109 X 2 X 10-6 / 0.552

E1 = 59504.132 i N/C

E2 = - 0.5 ( i + 1.73 j ) k q2 / L2

= - 0.5 X 9 X 109 X 7 X 10-6 ( i + 1.73 j ) / 0.552

= - 104132 X ( i + 1.73 j ) N/C

E2 = - ( 104132 i + 180148 j ) N/C

E = E1 + E2

E = 59504.132 i N/C - ( 104132 i + 180148 j ) N/C

E = - 44628 i - 180148 j N/C

E = - ( 44.628 i + 180.148 j ) kN/C

b )

F = qE

F = 2 X10-6 X ( - 44628 i - 180148 j )

F = ( - 0.089256 i - 0.360296 j ) N

F = - ( 89.256 i + 360.296 j ) mN

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