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1) In the diagram above, charge q 2 is 2x10 -9 C and charge q 1 has mass 0.12g.

ID: 2270933 • Letter: 1

Question


1) In the diagram above, charge q2 is 2x10-9C and charge q1 has mass 0.12g. The separation r is 5.3cm, and the angle ? is 14.5 degrees. Find q1 (magnitude and sign, you don't need to enter a + for positive answers but you will need to enter a - if negative). Note: the answer for this question is very small and using scientific notation will help. This system uses E in place of x10 so for example to enter 3x10-24 type 3E-24 in the answer box.
In the diagram above, charge q2 is 2x10-9C and charge q1 has mass 0.12g. The separation r is 5.3cm, and the angle ? is 14.5 degrees. Find q1 (magnitude and sign, you don't need to enter a + for positive answers but you will need to enter a - if negative). Note: the answer for this question is very small and using scientific notation will help. This system uses E in place of x10 so for example to enter 3x10-24 type 3E-24 in the answer box. In the diagram above if q1 has mass 0.2g and charge q2 is 2.7x10-9C. The separation r is 4.2cm, and the angle ? is 15.2 degrees. Find the magnitude of the electric field produced by q2 at the location of q1. Please be sure to express your answer in N/C (you don't need to enter units with your answer). There are a few formulas listed on our lab manuel and several more in our book, but I cannot find a single one that incorporates all parts of this problem. This is our first week of lab and our second week of classes, so I am lost as to where this information/knowledge shoudl be coming from. Hope someone can help me.

Explanation / Answer

1)
let Fe is the electric force

and let T is the tension in the string.

Fnety = 0

T*cos(theta) - m*g = 0

T = m*g/cos(theta)

= 0.12*10^-3*9.8/cos(14.5)

= 1.215*10^-3 N

Fnetx = 0

Fe - T*sin(theta) = 0

Fe = T*sin(14.5)

= 1.215*10^-3*sin(14.5)

= 0.304*10^-3 N

Fe = k*q1*q2/r^2

q1 = Fe*r^2/k*q2

= 0.304*10^-3*0.053^2/(9*10^9*2*10^-9)

= 4.75*10^-8 C

= 47.5*10^-9 C