Dipole: Science is indeed mysterious and doesn’t stop to surprise you as and when different topics are discussed. We are already familiar with the fact that charge exists around us and its presence leads to several natural phenomena. In addition, positive and negative charges are present in different forms that showcase diverse properties in the attendance of a motivating field.
Have you ever heard about the concept of an electric dipole? This unique setup of electric charges, i.e., positive & negative charges does form an interesting concept of physics. To be precise, an Electric Dipole can be tagged as a separation between positive and negative charges.
For example, you can consider a pair of electric charges having opposite sign but equal magnitude, parted by a significantly smaller distance. Our focus at present is to analyze the behavior of an Electric Dipole in the presence of an external field. This information is carefully decoded and presented in the following sections.
Dipole Placed in Uniform External Field
Since the impact of an external electric field on charges is already known to us; a dipole too will experience some form of force when introduced to an external field. It is interesting to learn that, a dipole placed in an external electric field acquires a rotating effect. This rotating effect is termed as ‘torque’ felt by the dipole. Excitingly, the net torque can be calculated on the opposite charges present in a dipole for estimating the overall rotation.
Browse more Topics under Electric Charges And Fields
- Conductors and Insulators
- Electric Charge
- Basic Properties of Electric Charge
- Coulomb’s Law
- Electric Field
- Electric Field Lines
- Gauss’s Law
- Applications of Gauss’s Law
- Electric Flux
- Electric Dipole
Torque & Its Calculation
In order to find torque experienced by a dipole when placed in an external field, let us consider that the dipole is introduced to a uniform external field. This field will generate an electric force having qE magnitude on the positive charge in the upward direction, whereas downward direction for the negative charge.
We can spot that the dipole rests in transitional equilibrium since the net force is zero. But what is the rotational equilibrium? Considering this case, the dipole might stay in a stationary position but does rotate with a particular angular velocity.
This fact has been proven experimentally and reveals that both electrostatic forces (qE) function as a torque being applied in the clockwise direction. Therefore, the dipole does get to rotate when placed in the uniform external electric field.
Always remember that torque always operates in a couple. Moreover, its magnitude is equivalent to the resultant product of force & its arm. Here, the arm can be seen as the distance falling between the point at which force operates and the point at which rotation occurs for the dipole.
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Dipole Placed in the Uniform External Electric Field
For a dipole placed in the uniform external electric field, the origin is taken as the point. Furthermore, torque is represented by ‘τ’ symbol. Do remember that, torque is a vector quantity. As per mathematical basis,
Torque magnitude (τ) = q E × 2a sin θ
τ = 2 q a E sin θ
τ = p E sin θ (Since p = 2 q a)
Solved Examples for You
Question: Give a real-life example of a dipole and electric field.
Solution: Try combing your dry hair and quickly bring the comb to several paper pieces. It would be observed that the comb pulls the paper pieces. This is because the comb acquires charge due to induction.
In a different way, the comb is known to polarize the paper pieces, i.e., produce a net dipole moment (direction of the electric field). Further, since the electric field stays non-uniform, the pieces of paper get attracted in the direction of the comb.
Question: An electric dipole is placed at an angle of 30o with an electric field of intensity 2×105 N/C. It experiences a torque of 4Nm. Calculate the charge on the dipole if the dipole length is 2cm.
Solution: B. Torque τ = p × E = pEsinθ
4 = p × 2 × 105 × sin30°
Or, p = 42 × 105 × sin30° = 4 × 10 −5 Cm
q = p / l = 4 × 1 0−5 / 0.02 = 2 × 10−3 C =2 mC