Find the Electric field strength at a point between two charges

[rocapp]

Homework Statement

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Homework Equations

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The Attempt at a Solution

I have tried parts C and D several times, and the homework program is telling me that it's still not correct. For C, I keep getting:

7.7x10^8 N/C

for D I keep getting:

7.1x10^-12 N/C

In order to find these, I know you just plug in the known values into the equation from part A. But each time I plug these in, I get an incorrect answer. Please help. Thanks in advance!

 

[SammyS]

Homework Statement

See images

Homework Equations

See images

The Attempt at a Solution

I have tried parts C and D several times, and the homework program is telling me that it's still not correct. For C, I keep getting:

7.7x10^8 N/C

for D I keep getting:

7.1x10^-12 N/C

In order to find these, I know you just plug in the known values into the equation from part A. But each time I plug these in, I get an incorrect answer. Please help. Thanks in advance!

For parts C & D please show your work, i.e. show exactly what numbers you are putting into your formula.

You are using millimeters and nanoCoulombs, aren't you ?

 

[rocapp]

Oh! Units were the problem. Does this change the value of the permittivity constant from 8.85x10^-12 F/m to 8.85x10^-9 F/mm?

 

[rocapp]

That is correct. I fixed the problems. Thanks!

 

[Nickie]

I would first check my calculations to ensure that I am using the correct equations and plugging in the values correctly. I would also double check the units and make sure they are consistent throughout the calculation. Additionally, I would check if there are any other factors that could be affecting the electric field strength at the point between the two charges, such as the presence of other charges or the distance between the charges. If all of these factors have been taken into account and the calculated values still do not match the expected answer, I would consider seeking assistance from a colleague or consulting a textbook or other reliable source. It is important to ensure the accuracy of calculations in the field of science, so I would not simply accept the incorrect answer and would continue to troubleshoot until the correct solution is found.