Potential at a point inside a solid non conducting uniformly charged sphere

[Rhdjfgjgj]

Homework Statement

Today I was deriving the potential of a point inside a solid non conducting sphere with uniform charge density. Let me show how my sir did this. For that please look at the following pic



As you can see my sir took the integral from r to R.(r is a point's distance from centre). And I chose to take integral limits from R to r. My answer is different For my calculations look below .

Can anyone tell why has this gone wrong.

Relevant Equations

E=KQ/r^2 , r>R
E=(KQ/R^3)r

My friend said 'you missed the negative sign the second time you wrote integral of Edr in the right hand side in your second line'. But if put a negative sign it would mean that potential decreases when travelling against ( antiparallel) to field, which is wrong

 

[TSny]

My friend said 'you missed the negative sign the second time you wrote integral of Edr in the right hand side in your second line'.

Yes, your friend is right. You should have a negative sign in the second line so that $\Delta V = -\int_R^r E dr$. The overall expression on the right side is positive. Since you are integrating from larger $r$ to smaller $r$, $dr$ in the integral is negative. So, the integral $\int_R^r E dr$ represents the summation of a bunch of infinitesimal negative quantities. Thus, $\int_R^r E dr$ will yield a negative quantity. Therefore, $-\int_R^r E dr$ is positive.

Going back to the first line, you wrote correctly that $$V_r - V_S = -\int_R^r \vec E \cdot \vec {dr}.$$ Here $\vec {dr}$ denotes an infinitesimal displacement along the radial path connecting the initial and final points. When integrating radially inward from the surface of the sphere, $\vec{dr}$ points radially inward while $\vec E$ points radially outward. Using $\vec A \cdot \vec B = |\vec A||\vec B| \cos \theta$, the dot product in the integrand is $$\vec E \cdot \vec {dr} = |\vec E||\vec {dr}| \cos \pi = -|\vec E||\vec {dr}|.$$ Here, we used $\theta = \pi$ since $\vec {dr}$ and $\vec E$ have opposite directions. $|\vec E|$ may be written simply as $E$, where $E$ denotes the magnitude of the electric field field vector $\vec E$. However, it is wrong to write $|\vec {dr}|$ as $dr$ since the notation $dr$ is reserved for the infinitesimal change in the radial distance $r$ when integrating from $R$ to $r$. Thus, $dr$ is a negative quantity when integrating inward. The correct relation for this case is $|\vec {dr}| = -dr$. So, putting all of this tediousness together, $$V_r - V_S = -\int_R^r \vec E \cdot \vec {dr} =+\int_R^r E |\vec {dr}| = -\int_R^r E dr.$$
It is less confusing if you use the rule that switching the limits of integration changes the sign of an integral. (This rule applies to line integrals as well "ordinary" integrals.) So, we have $$V_r - V_S = -\int_R^r \vec E \cdot \vec {dr} = +\int_r^R \vec E \cdot \vec {dr}. $$ Now we're integrating outward so that $\vec E$ and $\vec {dr}$ point in the same direction and $|\vec {dr}| = dr$. $$V_r - V_S =+\int_r^R E dr. $$

 

[Rhdjfgjgj]

Yes, your friend is right. You should have a negative sign in the second line so that $\Delta V = -\int_R^r E dr$. The overall expression on the right side is positive. Since you are integrating from larger $r$ to smaller $r$, $dr$ in the integral is negative. So, the integral $\int_R^r E dr$ represents the summation of a bunch of infinitesimal negative quantities. Thus, $\int_R^r E dr$ will yield a negative quantity. Therefore, $-\int_R^r E dr$ is positive.

Going back to the first line, you wrote correctly that $$V_r - V_S = -\int_R^r \vec E \cdot \vec {dr}.$$ Here $\vec {dr}$ denotes an infinitesimal displacement along the radial path connecting the initial and final points. When integrating radially inward from the surface of the sphere, $\vec{dr}$ points radially inward while $\vec E$ points radially outward. Using $\vec A \cdot \vec B = |\vec A||\vec B| \cos \theta$, the dot product in the integrand is $$\vec E \cdot \vec {dr} = |\vec E||\vec {dr}| \cos \pi = -|\vec E||\vec {dr}|.$$ Here, we used $\theta = \pi$ since $\vec {dr}$ and $\vec E$ have opposite directions. $|\vec E|$ may be written simply as $E$, where $E$ denotes the magnitude of the electric field field vector $\vec E$. However, it is wrong to write $|\vec {dr}|$ as $dr$ since the notation $dr$ is reserved for the infinitesimal change in the radial distance $r$ when integrating from $R$ to $r$. Thus, $dr$ is a negative quantity when integrating inward. The correct relation for this case is $|\vec {dr}| = -dr$. So, putting all of this tediousness together, $$V_r - V_S = -\int_R^r \vec E \cdot \vec {dr} =+\int_R^r E |\vec {dr}| = -\int_R^r E dr.$$
It is less confusing if you use the rule that switching the limits of integration changes the sign of an integral. (This rule applies to line integrals as well "ordinary" integrals.) So, we have $$V_r - V_S = -\int_R^r \vec E \cdot \vec {dr} = +\int_r^R \vec E \cdot \vec {dr}. $$ Now we're integrating outward so that $\vec E$ and $\vec {dr}$ point in the same direction and $|\vec {dr}| = dr$. $$V_r - V_S =+\int_r^R E dr. $$

So what you are saying is that when integrating against r (like moving radially in , we must consider '-dr' and not 'dr'.
Is it also applicable in case of a point change. Can u just show me. We will try by finding V at a general r , knowing that V at infinity=0. Can you please do the process for me. Thanks

 

[Rhdjfgjgj]

Moreover can anyone who's answering it tell me which concept is wrong for me to have gotten a doubt like this.

 

[kuruman]

Moreover can anyone who's answering it tell me which concept is wrong for me to have gotten a doubt like this.

If anything, you don't have a consistent way to do line integrals. The method shown below is straightforward, because it asks you to do what the line integral says you ought to do. You write the vectors in unit vector notation and then let the limits of integration take care of the negative signs if any. Here is what I mean using as example this problem in the case $r>R.$

Step 1. Write the vectors in unit vector notation (I use bold characters for vectors)
$\mathbf{ E}=\dfrac{kQ}{r^2}\hat{\mathbf{r}}~$; $~d\mathbf{r}=dr\hat{\mathbf{r}}.$

Step 2.
Take the dot product
$\mathbf{ E}\cdot d\mathbf{r}=\dfrac{kQ}{r^2}dr(\hat{\mathbf{r}}\cdot\hat{\mathbf{r}})=\dfrac{kQ}{r^2}dr.$

Step 3.
Assemble the integral noting that the lower limit is the starting point and the upper limit the ending point.
$$V(R)-V(r)=-\int_r^R \mathbf{ E}\cdot d\mathbf{r}=-\int_r^R\frac{kQ}{r^2}dr.$$ Note that vector $~d\mathbf{r}$ by definition points in the direction of increasing $r$ regardless of the sense of integration. The limits of integration automatically take care of relative negative signs. If the charge on the sphere were negative, you would just write $\mathbf{ E}=-\dfrac{kQ}{r^2}\hat{\mathbf{r}}$ and proceed as before.