1)(3 points) Using the same figure as in the previous problem, find the magnitud

Question

1)(3 points) Using the same figure as in the previous problem, find the magnitude of the total Coulomb force on the charge q in the center, given that charge q = 1.00x10^6 C, charge a = 2.00x10^6 C, charge b = -3.00x10^6 C, charge c = -4.00x10^6 C, and charge d = 1.00x10^6 C . The square is now 50.0 cm on a side. All the charges in the corners are located at the corners of the square. (answer in N)

Find the magnitude of the electric field at the center of the triangular configuration of charges in the figure shown below, given that charge a is 2.50nC, charge b is negative 8.00nC, charge c is 1.50nC, and the length of each side of the triangle is 25.0cm. Note the triangle is an equilateral triangle. Finding the center is not trivial, I suggest doing some research; Wikipedia is a good place to start. Note the number of significant figures in the problem! Just like all other calculations in this homework, keep extra digits in all calculations until the very end result is obtained. (answer in x1033 N/C. )

(1 Point) Find the direction of the electric field at the center of the triangular charge configuration in the previous problem. Report the number value in degrees. Note an angle measure above the x-axis corresponds to a positive number an angle below the x axis is a negative number. The reported angle should be measured from the positive x-axis. Again, note the number of significant figures in the problem. Just like all other calculations in this homework and future ones, keep extra digits in all calculations until the very end result is obtained.

Explanation / Answer

r = distance from center to corner = d/sqrt2 = 50 /sqrt2 = 0.5/sqrt2

Fax = k*qa*q*cos45/r^2 = (9*10^9*2*10^-6*1*10^-6cos45*2)/0.5^2 = 0.101 N   

Fay = -k*qa*q*sin45/r^2 = -(9*10^9*2*10^-6*1*10^-6sin45*2)/0.5^2 = -0.101 N

Fbx = k*qb*q*cos45/r^2 = (9*10^9*3*10^-6*1*10^-6cos45*2)/0.5^2 = 0.153 N

Fby = k*qb*q*sin45/r^2 = -(9*10^9*3*10^-6*1*10^-6sin45*2)/0.5^2 = 0.153 N

Fcx = -k*qc*q*cos45/r^2 = -(9*10^9*4*10^-6*1*10^-6cos45*2)/0.5^2 = -0.204 N

Fcy = -k*qc*q*sin45/r^2 = -(9*10^9*4*10^-6*1*10^-6sin45*2)/0.5^2 = -0.204 N

Fdx = -k*qd*q*cos45/r^2 = -(9*10^9*1*10^-6*1*10^-6cos45*2)/0.5^2 = -0.051 N

Fdy = -k*qd*q*sin45/r^2 = (9*10^9*1*10^-6*1*10^-6sin45*2)/0.5^2 = 0.051 N

Fx = 0.101 + 0.153 - 0.204 - 0.051 = -0.001 N

Fy = -0.101 + 0.153 - 0.204 + 0.051 = -0.101 N

F = sqrt(Fx^2+Fy^2) = sqrt(0.001^2+0.101^2) = 0.101 N

+++++++++++++++++++++

r = d/sqrt3 = 25/sqrt3 = 14.43 cm = 0.1443 m

Eax = 0

Eay = -k*qa/r^2 = -(9*10^9*2.5*10^-9)/0.1443^2 = -1080.56 N/C

Ebx = k*qb*cos30/r^2 = (9*10^9*8*10^-9*cos30)/0.1443^2 = 2994.54 N/C

Eby = - k*qb*sin30/r^2 = -(9*10^9*8*10^-9*sin30)/0.1443^2 = 1728.8 N/C

Ecx = k*qc*cos30/r^2 = (9*10^9*1.5*10^-9*cos30)/0.1443^2 = 561.5 N/C

Ecy = k*qc*sin30/r^2 = (9*10^9*1.5*10^-9*sin30)/0.1443^2 = 324.17 N/C

Ex = 0 + 2994.54 + 561.5 = 3556.04 N/C

Ey = -1080.56 + 1728.8 + 324.17 = 972.41 N/C

E = sqrt(3556.04^2+972.41^2) = 3686.6 N/C = 3.69*10^3 N/C

direction = tan^-1(972.41/3556.04) = + 15.3

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.