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]

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

(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]

(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

hey ! what formula did u use to calculate in part a??

[Deadly_king]

hey ! what formula did u use to calculate in part a??

The formula of electric potential from the formulae sheet.

Electric potential = Q / (4 x 22/7 x E_or)
where Q =  Charge
E_o, Epsilon note = 8.85 x 10^-12
r = separation distance 

[Assi1993]

The formula of electric potential from the formulae sheet.

Electric potential = Q / (4 x 22/7 x E_or)
where Q =  Charge
E_o, Epsilon note = 8.85 x 10^-12
r = separation distance 

! thanks! !!!!!!!!many many prayrs for u.. !!

[Deadly_king]

! thanks! !!!!!!!!many many prayrs for u.. !!

You're welcome

Ooh.....that's very kind of you.......thank you