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electric fields question - point charges

An atom is being ionised. The final electron starts at a distance of 2.4 x 10^-10m from the nucleus. The charges on the electron and the nucleus are 1.6 x 10^-19C and 8.0 x 10^-19 respectively.
a.) calculate the force between the electron and the nucleus.

i can do this part and i get 2.0 x 10^-8N - cant do next bit tho - dont undertsand it:

b.) by using the inverse square law, deduce values of force at distances of 4.8 x 10^-10m, 9.6 x 10^-10m and 19.2 x 10^-10m.

sum1 help pls - anything at all would be good.
ryan750
An atom is being ionised. The final electron starts at a distance of 2.4 x 10^-10m from the nucleus. The charges on the electron and the nucleus are 1.6 x 10^-19C and 8.0 x 10^-19 respectively.
a.) calculate the force between the electron and the nucleus.

i can do this part and i get 2.0 x 10^-8N - cant do next bit tho - dont undertsand it:

b.) by using the inverse square law, deduce values of force at distances of 4.8 x 10^-10m, 9.6 x 10^-10m and 19.2 x 10^-10m.

sum1 help pls - anything at all would be good.

Well the force is inversely proportional to the square of the distance-
F=kx1d2\mathrm{F = k x \frac{1}{d^2}}
So the force at double the distance is going to be 4 times smaller:
At 4.8.. = 1/4 x 2.0 x 10^-8N
At 9.6.. = 1/16 x 2.0 x 10^-8N
At 19.2.. = 1/64 x 2.0 x 10^-8N
Reply 2
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ryan750
OP
2
Golden Maverick
Well the force is inversely proportional to the square of the distance-
F=kx1d2\mathrm{F = k x \frac{1}{d^2}}
So the force at double the distance is going to be 4 times smaller:
At 4.8.. = 1/4 x 2.0 x 10^-8N
At 9.6.. = 1/16 x 2.0 x 10^-8N
At 19.2.. = 1/64 x 2.0 x 10^-8N


excellent thanks. i have a different question also that im struggling with;

An electric field is created by a point charge, Q, of 4.3 x 10^-8C. Point A is 0.25m from Q. Point B is 0.5m from Q.
A 1.7 x 10^-10C charge is placed at point B in the field described above.
a.) calculate the force acting on the smaller charge.

i can do this part and i get 2.6 x 10-7N

b.) calculate the work done moving the smaller charge to B from a position outside the field.

i cant do part b - the answer in the book is 1.3 x 10^-7.
ryan750
excellent thanks. i have a different question also that im struggling with;

An electric field is created by a point charge, Q, of 4.3 x 10^-8C. Point A is 0.25m from Q. Point B is 0.5m from Q.
A 1.7 x 10^-10C charge is placed at point B in the field described above.
a.) calculate the force acting on the smaller charge.

i can do this part and i get 2.6 x 10-7N

b.) calculate the work done moving the smaller charge to B from a position outside the field.

i cant do part b - the answer in the book is 1.3 x 10^-7.

This isn't in my syllabus and so I don't know how to explain it properly

W = qV

Electric potential due to Q:
V = Q4πϵ0r \Large V \ = \ \frac{Q}{4\pi\epsilon_0 r}
so
Unparseable latex formula:

\Large W \ = \ \frac{qQ}{4\pi\epsilon_0 r} \\ \, \\[br]W \ = \ \frac{1.7*10^{-10}*4.3*10^{-8}}{4\pi\8.85*10^{-12}*0.5}

Reply 4
Avatar for ryan750
ryan750
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2
thanks - but i just find the wording of the question really weird. wot does it mean 'moving the smaller charge to B' the smallest charge is already at b. And then moving it from a position outside the field? this question is strange.

Any interpretations or explanations welcome :tsr:
Reply 5
Avatar for habosh
habosh
13
ryan750
excellent thanks. i have a different question also that im struggling with;

An electric field is created by a point charge, Q, of 4.3 x 10^-8C. Point A is 0.25m from Q. Point B is 0.5m from Q.
A 1.7 x 10^-10C charge is placed at point B in the field described above.
a.) calculate the force acting on the smaller charge.

i can do this part and i get 2.6 x 10-7N

b.) calculate the work done moving the smaller charge to B from a position outside the field.

i cant do part b - the answer in the book is 1.3 x 10^-7.

the work done equals to the potential energy of the field,like if you remember the energy level of the electrons where out side the atom the energy is zero and as you get further you get negative energies,so the enrgy needed for the charge to escape from B is the work done moving the charge to B(conversation of energy)
now FX=W
you found the force and it represents mg in PE=mgh
so x is the distancebetween B and the nucleus is .5m
simply 2.6 x10^-7 X.5=1.3 x 10^-10
I hope you get it now:wink:
Reply 6
Avatar for habosh
habosh
13
ryan750
thanks - but i just find the wording of the question really weird. wot does it mean 'moving the smaller charge to B' the smallest charge is already at b. And then moving it from a position outside the field? this question is strange.

Any interpretations or explanations welcome :tsr:

well there are two charges the point charge and the one at B so they moved the smaller one closer from the outside to B
Reply 7
Avatar for ryan750
ryan750
OP
2
very clever - i get it now - thanks so much.

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