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In the problem I have been told the value for:

Fe = 5.5 x 10-9, r = 5.0 x 10 -10 and of course Ke (8.99 x 10 9)

known nucleus charge is e= 1.6 x 10-19.

I have been asked whether the magnitude of known nuclues 1, and unknown nuclus is attractive or repulsive and I reckon its repulsive since the product of q1 and q2 has been given as a positive (5.5 x 10-9)

The thing that has been driving me mad is how to calculate the electric charge on the unknown nucleus. The only thing I can think of is to use my rearranged equation (q = Fe X r squared, divided by Ke), and then take the answer, which I have worked out to be q = 1.528 x 10 -37 to calulate q2. Because if the sum of q1 and q2 is 1.528 x 10 -37 , then q2 must equal 1.528 x 10 -37 - 1.6 x 10-19 (known nucleus charge) = -1.6 x 10*19.

Is this right? and does this make it then an attractive force? Also is it right that both nuclei have same charge, one negative and one positive (I guess electrons and protons have the same charges...)

Sorry this is so long, but I have got myself in a complete muddle.

Any help greatly appreciated

Thank you

Jill