Question

ABC is a right angle triangle with sides ab is equals to 0.3 metre BC is equals to 0.4 metre and angle b is equal to 90 degree point charges + 18 Newton koolom and + 32 Newton koolom at placed at the corners and C respectively find the force acting on a charge to 2Mue placed at c

Answer 1

Given:

Charge at A, =qa 15 esu

Charge at B = qb12 esu

Charge at C, =qc -12 esu

Distance between A and B, r₁ = AB = 3 cm

Distance between B and C, r₂ = BC = 4 cm

Now,

the Force on B due to charge at A = kqkqb/r1/2

here,

k is the coulomb's constant = 1 for CGS system

therefore,

F₁ = 1*15*12/(3)2

or

F₁ = 20 dyne

similarly

the Force on B due to charge at C =

here,

k is the coulomb's constant = 1 for CGS system

therefore,

F₂ = 1*(-12)*12/(4)2

or

F₂ = -9 dyne

Now,

the resultant force on B due to both the charges is given as:

F = √f1/2+f2/2

or

F = √(20)2+(-9)2

or

F = 21.93 dyne ≈ 22 dyne

√\_____Anushka♥