Two negative and two positive point charges (magnitude Q = 4.16 mC ) are placed

Question

Two negative and two positive point charges (magnitude Q = 4.16 mC ) are placed on opposite corners of a square as shown in the figure (a = 0.135 m ).(Figure 1)

Part A

Determine the magnitude of the force on each charge.

Part B

Determine the direction of the force on each charge.

Please help!

Two negative and two positive point charges (magnitude Q = 4.16 mC ) are placed on opposite corners of a square as shown in the figure (a = 0.135 m ).(Figure 1)

Part A

Determine the magnitude of the force on each charge.

Part B

Determine the direction of the force on each charge.

Please help!

Phys 62 HW 8: Electric Forces Problem 21.16 X Incorrect; Try Again; 7 attempts remaining Part B 2 Determine the direction of the force on each charge Enter your answers numerically separated by commas 4 0,02,03,04 counterclockwise from the +x axis

Explanation / Answer

r^2 = a^2+a^2 = 2a^2 = 2*0.135^2 = 0.03645

for charge 1

F2x = k*Q^2/a^2

F4y = -k*Q^2/a^2

F3x = -k*Q^2*cos45/r^2 = -k*Q^2*cos45/(2a^2)

F3y = k*Q^2*sin45/r^2 = k*Q^2*sin45/(2a^2)

F1x = F2x + F3x = (k*Q^2/a^2)*(1 - cos45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(1-cos45/2) = 5.524*10^6 N

F1y = F4y + F3y = (k*Q^2/a^2)*(-1 + sin45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(-1+sin45/2) =-5.524*10^6 N

F1 = sqrt(F1x^2+F1y^2) = 7.81*10^6 N

direction = tan^-1(F1y/F1x) =315

-------------------------------------

for charge 2

F1x = -k*Q^2/a^2

F3y = -k*Q^2/a^2

F4x = k*Q^2*cos45/r^2 = k*Q^2*cos45/(2a^2)

F4y = k*Q^2*sin45/r^2 = k*Q^2*sin45/(2a^2)

F2x = F1x + F4x = (k*Q^2/a^2)*(-1 + cos45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(-1+cos45/2) = -5.524*10^6 N

F2y = F3y + F4y = (k*Q^2/a^2)*(-1 + sin45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(-1+sin45/2) = -5.524*10^6 N

F2 = sqrt(F2x^2+F2y^2) = 7.81*10^6 N

direction = tan^-1(F1y/F1x) = 225

----------------------------------------------------------------------

for charge 3

F4x = -k*Q^2/a^2

F2y = k*Q^2/a^2

F1x = k*Q^2*cos45/r^2 = k*Q^2*cos45/(2a^2)

F1y = -k*Q^2*sin45/r^2 = -k*Q^2*sin45/(2a^2)

F3x = F4x + F1x = (k*Q^2/a^2)*(-1 + cos45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(-1+cos45/2) = -5.524*10^6 N

F3y = F2y + F1y = (k*Q^2/a^2)*(1 - sin45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(1-sin45/2) = 5.524*10^6 N

F3 = sqrt(F2x^2+F2y^2) = 7.81*10^6 N

direction = tan^-1(F1y/F1x) = 135

---------------------------------------

for charge 4

F3x = k*Q^2/a^2

F1y = k*Q^2/a^2

F2x = -k*Q^2*cos45/r^2 = -k*Q^2*cos45/(2a^2)

F2y = -k*Q^2*sin45/r^2 = -k*Q^2*sin45/(2a^2)

F4x = F3x + F2x = (k*Q^2/a^2)*(1 - cos45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(1-cos45/2) = 5.524*10^6 N

F4y = F1y + F2y = (k*Q^2/a^2)*(1 - sin45/2) = (9*10^9*(4.16*10^-3)^2/0.135^2)*(1-sin45/2) = 5.524*10^6 N

F2 = sqrt(F2x^2+F2y^2) = 7.81*10^6 N

direction = tan^-1(F1y/F1x) = 45

part(A)

7.81*10^6 N , 7.81*10^6 N , 7.81*10^6 N , 7.81*10^6 N

part(B)

45 , 225 , 135 , 45

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