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Sahil Muhammed
Sahil Muhammed
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While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, $dq$ denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional continuous charge distribution)?

After some thought, it appears to me that maybe it is in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, $dq$ denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional charge distribution)?

After some thought, it appears to me that maybe it in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, $dq$ denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional continuous charge distribution)?

After some thought, it appears to me that maybe it is in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

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Qmechanic ♦
Qmechanic ♦
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While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, dq$dq$ denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional charge distribution)?

After some thought, it appears to me that maybe it in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, dq denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional charge distribution)?

After some thought, it appears to me that maybe it in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, $dq$ denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional charge distribution)?

After some thought, it appears to me that maybe it in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?

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Sahil Muhammed
Sahil Muhammed
  • 37
  • 7

Is it valid to write $\int{dq}$?

While I was going through the expression for the electric field in continuous charge distributions (Section 2.1.4) in Griffiths' textbook, it was given as $${E(r)} = \frac{1}{4\pi\epsilon_o}\int{\frac{1}{r^2}\hat{r}dq}$$

So, would it be valid to write this expression because the quantization of charge implies that there exists a lower limit to the possible value that charge can take (i. e. the charge of an electron (or proton, if you consider the magnitude), which is about $1.6$ $\cdot{}$ $10^{-19}$ C). However, dq denotes an infinitesimal piece of charge that we're using to integrate.

Also, is this why $dq$ is often computed by other means like using $\lambda$ (charge per unit length, in case of a one dimensional charge distribution)?

After some thought, it appears to me that maybe it in indeed incorrect to use the expression above and rather use $\Delta{q}$ instead and proceed to calculate the electric field by using Riemann sums instead. Quite tedious, but that's the only way I see.

Can you help me understand this and let me know if there is/are some flaws in my understanding?