two point charges q1=4uC q2=-9uC are placed so that the distance between them is 80 cm
find a) the point at which the electric field intensity is zero and find the electric potential at this point
b)the point at which the electric potential is zero and find the electric field intensity at this point
find a) the point at which the electric field intensity is zero and find the electric potential at this point
b)the point at which the electric potential is zero and find the electric field intensity at this point
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a) Suppose q1 is on the left and q2 is on the right.
q1---0.8m----q2
By symmetry the null point (zero resultant field) must lie somewhere along the line passing through q1 and q2. We need to find in which region the null point is.
It can't be between q1 and q2, because in this region the field from q1acts right (away from the +ve q1) and the field from q2 also acts right (towards the -ve q2) - the 2 fields could never cancel as they both act to the right.
It can't be to the right of q2. because in this region the field from q1 acts right and the field from q2 acts left BUT because |q2|>|q1| the 2 fields can never cancel as the field from q2 will always have a bigger magnitude than the field form q1.
It is to the left of q1. In this region the field from q1 acts left and the field from q2 acts right; since |q2>|q1| - the field from q2 can be big enough to cancel the field from q1.
Suppose the null point, P, is a distance x to the left of q1:
P--x---q1---0.8m----q2
Using E = kq/d², the magnitudes of the 2 fields must be equal at P, so:
k|q1|/x² =k|q2|/(x+0.8)²
4/x² =9/(x+0.8)² (the 10^-6 and k cancel)
2/x = 3/(x+0.8)
2x+1.6 = 3x
x = 1.6m
The field intensity is zero along the line joining q1 and q2, 1.6m to the outside of q1.
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b) This is easier as the potentials simply add (they are scalars). At some point between q1 an q2 the positive and negative potentials add to zero. If the point is a distance x from q1:
q1---x---P---q2
Using V = kq/d
kq1/x + kq2/(0.8-x) =0
4/x + (-9)/(0.8-x) =0
4/x = 9/(0.8-x)
3.2 - 4x = 9x
x = 3.2/13 = 0.25m
The potential is zero along the line joining q1 and q2, 0.25m from q1, in the region between q1 and q2.
q1---0.8m----q2
By symmetry the null point (zero resultant field) must lie somewhere along the line passing through q1 and q2. We need to find in which region the null point is.
It can't be between q1 and q2, because in this region the field from q1acts right (away from the +ve q1) and the field from q2 also acts right (towards the -ve q2) - the 2 fields could never cancel as they both act to the right.
It can't be to the right of q2. because in this region the field from q1 acts right and the field from q2 acts left BUT because |q2|>|q1| the 2 fields can never cancel as the field from q2 will always have a bigger magnitude than the field form q1.
It is to the left of q1. In this region the field from q1 acts left and the field from q2 acts right; since |q2>|q1| - the field from q2 can be big enough to cancel the field from q1.
Suppose the null point, P, is a distance x to the left of q1:
P--x---q1---0.8m----q2
Using E = kq/d², the magnitudes of the 2 fields must be equal at P, so:
k|q1|/x² =k|q2|/(x+0.8)²
4/x² =9/(x+0.8)² (the 10^-6 and k cancel)
2/x = 3/(x+0.8)
2x+1.6 = 3x
x = 1.6m
The field intensity is zero along the line joining q1 and q2, 1.6m to the outside of q1.
_________________________________
b) This is easier as the potentials simply add (they are scalars). At some point between q1 an q2 the positive and negative potentials add to zero. If the point is a distance x from q1:
q1---x---P---q2
Using V = kq/d
kq1/x + kq2/(0.8-x) =0
4/x + (-9)/(0.8-x) =0
4/x = 9/(0.8-x)
3.2 - 4x = 9x
x = 3.2/13 = 0.25m
The potential is zero along the line joining q1 and q2, 0.25m from q1, in the region between q1 and q2.
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E equals mc squared