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electric field
winnie101:
how do i solve the following
2 opposite charges of the same magnitude 2*10^-7 C are separated by a distance of 15cm.
a) what is the magnitude and direction of the electric field strength at the midpoint between the charges?
b)what is the magnitude and direction of the force acting on an electron placed at the midpoint?
Thanks
Deadly_king:
--- Quote from: winnie101 on October 08, 2010, 05:24:00 pm ---how do i solve the following
2 opposite charges of the same magnitude 2*10^-7 C are separated by a distance of 15cm.
a) what is the magnitude and direction of the electric field strength at the midpoint between the charges?
b)what is the magnitude and direction of the force acting on an electron placed at the midpoint?
Thanks
--- End quote ---
(a) The resultant electric field at a point due to a system of point charges is given by the vector sum of the electric field intensities due to the individual charges.
Let Q1 = 2x10-7 C and Q2 = -2x10-7 C
Electric field strength due to Q1 = (2x10-7)/(4 x 22/7 x (8.85x10-12) x (15x10-2)/2) = 2.40 x 104 Vm-1
Electric field strength due to Q2 will have same magnitude and in same direction since they are oppositely charged.
Resultant electric field intensity will be 2(2.40x104) = 4.80x104 Vm-1
The direction of the electric field is always along the direction of decreasing electric potential. Therefore from the positive charge to the negative one.
(b) Use the formula F = qV/d
Force = (2.0x10-7)x(4.80x104)/((15x10-2)/2) = 6.39x10-2 N
The direction of the force towards the positive charge, that is against the direction of the electric field.
Hope it helps :)
Assi1993:
--- Quote from: Deadly_king on October 09, 2010, 06:09:10 am ---(a) The resultant electric field at a point due to a system of point charges is given by the vector sum of the electric field intensities due to the individual charges.
Let Q1 = 2x10-7 C and Q2 = -2x10-7 C
Electric field strength due to Q1 = (2x10-7)/(4 x 22/7 x (8.85x10-12) x (15x10-2)/2) = 2.40 x 104 Vm-1
Electric field strength due to Q2 will have same magnitude and in same direction since they are oppositely charged.
Resultant electric field intensity will be 2(2.40x104) = 4.80x104 Vm-1
The direction of the electric field is always along the direction of decreasing electric potential. Therefore from the positive charge to the negative one.
(b) Use the formula F = qV/d
Force = (2.0x10-7)x(4.80x104)/((15x10-2)/2) = 6.39x10-2 N
The direction of the force towards the positive charge, that is against the direction of the electric field.
Hope it helps :)
--- End quote ---
hey ! what formula did u use to calculate in part a??
Deadly_king:
--- Quote from: Assi1993 on October 12, 2010, 08:00:35 am ---
hey ! what formula did u use to calculate in part a??
--- End quote ---
The formula of electric potential from the formulae sheet.
Electric potential = Q / (4 x 22/7 x Eor)
where Q = Charge
Eo, Epsilon note = 8.85 x 10-12
r = separation distance
Assi1993:
--- Quote from: Deadly_king on October 12, 2010, 08:14:23 am ---The formula of electric potential from the formulae sheet.
Electric potential = Q / (4 x 22/7 x Eor)
where Q = Charge
Eo, Epsilon note = 8.85 x 10-12
r = separation distance
--- End quote ---
! thanks! :) !!!!!!!!many many prayrs for u.. !!
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