Question

solve this question for me !!!???
With steps .

Answer 1

Heya!!

Here The charges +10^-8C and -10^-8C are held at P and Q respectively .
PQ = 0.1 m
PA = PB = 0.05 m
QA = 0.05m
CP = CQ = 0.1 m
At A , Field Intensity due to charge
--------------------------------------------------
q1 = Kq×1/AP²
= 9×10^9 × 10-^8/(0.05)²
= 36000N/C
Field intensity due to charge
q2 = 9×10^9×10^8/(0.05)1
= 36000N/C
Net field intensity at A = 36000+36000 = 72000N/C along AQ .


At B Field Intensity due to charge q1
= Kq/(PB)²
= 9×10^9 ×10^-8/(0.05)²
= 36000N/C.
Field Intensity due to charge q2
= 9×10^9 × 10^-8 / (0.15)² = 4000N/C

So, Net field intensity at B
= 36000-4000 = 32000N/C


At C Field Intensity due to charge q1

E1= k q/(PC)²
= 9×10^9 × 10^-8 /(0.1)²
= 9×10³N/C

Field Intensity due to charge q2
E2= Kq/QC²

= 9×10^9 ×10^-8 /(0.01)² = 9×10³ N/C
Both E1&E2 are equal in magnitude acting at 60° to CR parallel to BA .
:• Resultant intensity at R
= E1cos∅ + E2cos∅
Here ∅ = 60°
=> 2E1cos∅
=> 2×9×10³×1/2 = 9×10³N/C
The component sin∅ in opposite direction will be cancelled out.