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A thin glass rod of length 2L has a linear charge density that is zero in the middle of the length of the rod and increases linearly (both positively) along the length of the rod in both directions from the middle of the rod. The electric potential at a point P, a distance y above the middle of the rod (directly above where the charge density is zero), is V = 2keb(/1? – y? – y), where b is the charge density coefficient. а. From this expression for the voltage, determine the expression for the electric field at point P. E = b. Verify this expression for the electric field by using the formula for electric field (Coulomb's Law divided by the field charge) for each differential bit of charge and integrating over the rod. (I want to see all the details for setting up the problem, as well as for evaluating the integral.) i. Start by drawing a diagram, labeling a coordinate system, and considering arbitrary but symmetrically-located differential bits of the rod. Label the distances, r, from the source points (the differential bits of charge) to the field point P, as well as the resulting differential E-field vectors from the differential bits of charge. Finally use these vectors to show that certain components cancel while others add.

A thin glass rod of length 2L has a linear charge density that is zero in the middle of the length of the rod and increases linearly (both positively) along the length of the rod in both directions from the middle of the rod. The electric potential at a point P, a distance y above the middle of the rod (directly above where the charge density is zero), is V = 2keb(/1? – y? – y), where b is the charge density coefficient. а. From this expression for the voltage, determine the expression for the electric field at point P. E = b. Verify this expression for the electric field by using the formula for electric field (Coulomb's Law divided by the field charge) for each differential bit of charge and integrating over the rod. (I want to see all the details for setting up the problem, as well as for evaluating the integral.) i. Start by drawing a diagram, labeling a coordinate system, and considering arbitrary but symmetrically-located differential bits of the rod. Label the distances, r, from the source points (the differential bits of charge) to the field point P, as well as the resulting differential E-field vectors from the differential bits of charge. Finally use these vectors to show that certain components cancel while others add.

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A thin glass rod of length 2L has a linear charge density that is zero in the middle of the length of the rod and increases linearly (both positively) along the length of the rod in both directions from the middle of the rod. The electric potential at a point P, a distance y above the middle of the rod (directly above where the charge density is zero), is V = 2keb(/1? – y? – y), where b is the charge density coefficient. а. From this expression for the voltage, determine the expression for the electric field at point P. E = b. Verify this expression for the electric field by using the formula for electric field (Coulomb's Law divided by the field charge) for each differential bit of charge and integrating over the rod. (I want to see all the details for setting up the problem, as well as for evaluating the integral.) i. Start by drawing a diagram, labeling a coordinate system, and considering arbitrary but symmetrically-located differential bits of the rod. Label the distances, r, from the source points (the differential bits of charge) to the field point P, as well as the resulting differential E-field vectors from the differential bits of charge. Finally use these vectors to show that certain components cancel while others add.
Transcribed Image Text:A thin glass rod of length 2L has a linear charge density that is zero in the middle of the length of the rod and increases linearly (both positively) along the length of the rod in both directions from the middle of the rod. The electric potential at a point P, a distance y above the middle of the rod (directly above where the charge density is zero), is V = 2keb(/1? – y? – y), where b is the charge density coefficient. а. From this expression for the voltage, determine the expression for the electric field at point P. E = b. Verify this expression for the electric field by using the formula for electric field (Coulomb's Law divided by the field charge) for each differential bit of charge and integrating over the rod. (I want to see all the details for setting up the problem, as well as for evaluating the integral.) i. Start by drawing a diagram, labeling a coordinate system, and considering arbitrary but symmetrically-located differential bits of the rod. Label the distances, r, from the source points (the differential bits of charge) to the field point P, as well as the resulting differential E-field vectors from the differential bits of charge. Finally use these vectors to show that certain components cancel while others add.
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