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guoxu
guoxu
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If we have several point charges $q_1, q_2, q_3, \cdots$, and their distances from a test charge $Q$ are respectively ${\imath}_1, {\imath}_2, {\imath}3$${\imath}_1, {\imath}_2, {\imath}_3$, $\cdots$, then the electric field intensity at $Q$ is given by $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$

This equation can be written in integral form as: $E(r) = \frac{1}{4 \pi \varepsilon_0} \int \frac{1}{\imath^2} \hat{\imath} dq$.

I don't quite understand why, when converting from the summation form $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$ to the integral form, $q$ is chosen as the variable of integration. Both $q$ and ${\imath}$ have the subscript $i$, so why not choose ${\imath}$?

If we have several point charges $q_1, q_2, q_3, \cdots$, and their distances from a test charge $Q$ are respectively ${\imath}_1, {\imath}_2, {\imath}3$, $\cdots$, then the electric field intensity at $Q$ is given by $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$

This equation can be written in integral form as: $E(r) = \frac{1}{4 \pi \varepsilon_0} \int \frac{1}{\imath^2} \hat{\imath} dq$.

I don't quite understand why, when converting from the summation form $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$ to the integral form, $q$ is chosen as the variable of integration. Both $q$ and ${\imath}$ have the subscript $i$, so why not choose ${\imath}$?

If we have several point charges $q_1, q_2, q_3, \cdots$, and their distances from a test charge $Q$ are respectively ${\imath}_1, {\imath}_2, {\imath}_3$, $\cdots$, then the electric field intensity at $Q$ is given by $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$

This equation can be written in integral form as: $E(r) = \frac{1}{4 \pi \varepsilon_0} \int \frac{1}{\imath^2} \hat{\imath} dq$.

I don't quite understand why, when converting from the summation form $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$ to the integral form, $q$ is chosen as the variable of integration. Both $q$ and ${\imath}$ have the subscript $i$, so why not choose ${\imath}$?

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guoxu
guoxu
  • 77
  • 5

Why is charge $q$ chosen as the variable of integration when converting from summation to integral form in electric field calculations?

If we have several point charges $q_1, q_2, q_3, \cdots$, and their distances from a test charge $Q$ are respectively ${\imath}_1, {\imath}_2, {\imath}3$, $\cdots$, then the electric field intensity at $Q$ is given by $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$

This equation can be written in integral form as: $E(r) = \frac{1}{4 \pi \varepsilon_0} \int \frac{1}{\imath^2} \hat{\imath} dq$.

I don't quite understand why, when converting from the summation form $E(r) \equiv \frac{1}{4 \pi \varepsilon_0} \sum_{i=1}^n \frac{q_i}{{\imath}_i^2} \hat{{\imath}}_i$ to the integral form, $q$ is chosen as the variable of integration. Both $q$ and ${\imath}$ have the subscript $i$, so why not choose ${\imath}$?