Question

40
consider tive charges + 30 units and +60 units placed 12 cm apart in air. Find the field intensity at the midway between two charges.
pe
A
0.83 dyne towards +60 unit charge
1
o 1 dyne
G
0.83 dyne towards + 30 unit charge
D
O
0.75 dyne
- AntWhat is the nature or the one thing on the manner
and on time is riven by
AS​

Answer 1

Given :

Magnitude of charge 1 (q₁) = + 30 units

Magnitude of charge 2 (q₂) = + 60 units

Distance between the charges = 12cm

To Find :

The magnitude of electric field intensity midway between them

Solution :

• The electric field lines always moves away from the positive charges. The resultant field will be towards charge 1 as the magnitude of charge 2 is higher

• The electric field is given by the formula

             $E=\frac{kq}{r^{2} }$

• The resultant electric field due to two charges is

             $E=\frac{kq_{2} }{r^{2} }-\frac{kq_{1} }{r^{2}}$

             $E=\frac{k}{r^{2} } (q_{2} -q_{1} )$

             $E=\frac{9\times10^{9}\times(60-30)}{6^{2} }$

             $E = 0.75\:dyne$

The resultant electric field  midway between the charges is 0.75 dyne towards the 30 units charge.