Question

A pendulum of mass 80 milligram carrying a charge of 2 x 10-8 C is at rest in a horizontal
Uniform Electric field of 2 x 104 N/C. Find the tension in the thread of the pendulum and
the angle it makes With the vertical.​

Answer 1

Answer:

shown below with an appropriate eg:-

Explanation:

Mass =80mg

Mass =80mgCharge =2×10−8 Coulomb

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kg

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8C

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibrium

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we get

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE⇒tanθ=80×10−6×102×10−8×2×104

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE⇒tanθ=80×10−6×102×10−8×2×104tanθ=0.5

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE⇒tanθ=80×10−6×102×10−8×2×104tanθ=0.5θ=tan−1(0.5)

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE⇒tanθ=80×10−6×102×10−8×2×104tanθ=0.5θ=tan−1(0.5)θ=270

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE⇒tanθ=80×10−6×102×10−8×2×104tanθ=0.5θ=tan−1(0.5)θ=270Now,

Mass =80mgCharge =2×10−8 CoulombElectric field =20,000Vm−1Given, mass =80mg=8×10−6kgcharge =q=2×10−8CLet T be the tension in string and θ is the angle if string with vertical.In equilibriumTsinθ=qE⟶(1)Tcosθ=mg⟶(2)Divide (1) by (2) we gettanθ=mgqE⇒tanθ=80×10−6×102×10−8×2×104tanθ=0.5θ=tan−1(0.5)θ=270Now,tension, T=sinθqE=sin2702×10−8×2×104=8.8×10