Question

# A system has two charges ${q}_{A}=2.5×{10}^{-7}C$ and ${q}_{B}=-2.5×{10}^{-7}C$ is located at points $A\left(0,0,-15\right)cm$ and $B\left(0,0,+15cm\right)$ respectively. What is the total charge and electric dipole moment vector of the system?

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Solution

## Step 1. Given Data,${q}_{A}=2.5×{10}^{-7}C$ ${q}_{B}=-2.5×{10}^{-7}$$A\left(0,0,-15cm\right)andB\left(0,0,+15cm\right)$Step 2. Formula used,Dipole moment= $P=q×d$$d$ is the length of the dipole$q$ is the given chargeStep 3. Calculating the total charge,The amount of charge at A, ${q}_{A}=2.5×{10}^{-7}C$The amount of charge at B, ${q}_{B}=-2.5×{10}^{-7}$The total charge of the system is $q={q}_{A}+{q}_{B}=2.5×{10}^{-7}C=0$Hence the total charge is $0$.Step 4. Calculating the electric dipole momentDistance between points A and B,$d=15+15=30cm=0.3m$The electric dipole moment of the system,$P={q}_{A}×d=2.5×{10}^{-7}×0.3=7.5×{10}^{-8}cm$ along the positive z-axis.Hence, the electric dipole moment of the system is $7.5×{10}^{-8}$ along the positive z-axis.

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