Question

an electric dispole consist of two opposit charge of magnitude 1/3 into 10 minus 7 cal seprated by 2cm the dispole is placed is an external field of 3 into 10 to the power 7 N/Cal what maximum torque does the electric field exert on the dispole?(theta equal to 90 degree)​

Answer 1

Given data,Charges(q)=2×10

Given data,Charges(q)=2×10 −6

Given data,Charges(q)=2×10 −6 C,

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measured

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric field

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2 )×(2×10

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2 )×(2×10 5

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2 )×(2×10 5 )

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2 )×(2×10 5 )=12×10

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2 )×(2×10 5 )=12×10 −3

Given data,Charges(q)=2×10 −6 C,Distance(d)=3cm=3×10 −2 m electric field(E)=2×10 5 N/C.Torque(τ)=?We know that Torque(τ) to be measuredTorque=charge×distance×Electric fieldTorque(τ)=(2×10 −6 )×(3×10 −2 )×(2×10 5 )=12×10 −3 Nm