# Does any object placed in an electric field change the electric field?

Lets say I have a point charge of magnitude $+q$, All around it I would have a symmetric radial electric field. Now if I place a neutral object lets say a sphere (doesn't matter insulating or conducting) in this field some distane away from the point charge. A negative charge will be induced on the object near the point charge and a positive charge on the opposite side.

No matter how small this induced charge is, due to the radial distance of the two (positive and negative) there must be an increase/decrease in net electric field on either side of the object and mostly everywhere else too !

I hope that what I am thinking is wrong, because we have not been taught that anything placed in electric field would affect the field itself regardless of it's nature. But I can't figure out what am I thinking wrong, how to solve this dilemma ?

• It's evident that haven't solved Poissons equation. Try to read and electrodynamics book, where method of images and other techniques are developed to analyse the problem you proposed. But I guess that if you use really neutral matter, like Z bosons, there would be no change at all. – jinawee Jan 16 '14 at 12:34
• You are right, I haven't ! I have studied at high school level till now and very little of undergrad level. But still, how would you analyse and solve this dilemma ? When we are solving questions of electrostatics we frequently insert dielectrics and other objects in electric fields, when these get polarised, why don't they change the net field ? Or do they ? – Rijul Gupta Jan 16 '14 at 12:43
• They do. But in many cases you can make an approximation and suppose they don't. See: en.wikipedia.org/wiki/File:Electrostatic_induction.svg and en.wikipedia.org/wiki/Polarization_density – jinawee Jan 16 '14 at 12:51

If the material placed in the field of the positive charge is a conductor, the field will be distorted and the method to see the field is the image charges method. It will depend on the boundary conditions.

For a grounded conducting sphere

Field lines outside a grounded sphere for a charge placed outside a sphere.

For a non grounded conductor:

This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, until the electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive charge on the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized.

If the field is created by a point charge the geometry will change but the physics is the same.

If you have a positive point charge and bring into its field a dielectric, then the field lines will change again depending on constants as :

Figure 6.6.6 Electric field intensity in and around dielectric rod of Fig. 6.6.5 for (a) e_b > a and (b) e_b< e_a.

One can again imagine the geometric changes for a field from a sphere.

In summary, the field distorts with the presence of matter, differently for a conductor or dielectric .

• I have never been so happy to see an answer, thank you very much. It was driving me nuts! What they teach us in high school level is really really useless. – Rijul Gupta Jan 16 '14 at 15:31

I'm not quite sure I understand why you have a problem with this - every static charge is a source or a drain of the electric field, depending on its sign. So obviously the field of a single charge at the origin will be different from the field of three charges or any other configuration.

The electric potential of such an ensemble of pointlike charges $q_i$ at a specific point, measured by an uncharged observer, will be $$\Phi(\vec r) = \frac{1}{4 \pi \epsilon_0} \cdot \sum_i \frac{q_i}{\left| \vec r - \vec r_i \right|}$$ At this point, instead of evaluating the sum you can do a multipole expansion $$\Phi(\mathbf r) = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q}{r} + \frac{\mathbf r \cdot \mathbf p}{r^3} + \frac{1}{2} \sum_{k,l} Q_{kl} \frac{r_k \cdot r_l}{r^5}+... \right),$$which is essentially a Taylor series of ${\left| \vec r - \vec r_i \right|}^{-1}$. From there, you will get the electric field by taking the gradient of the approximated potential.

• You are saying " obviously the field of a single charge at the origin will be different from the field of three charges or any other configuration " But that is not what I am getting confused over. I am unable to comprehend how a neutral object can change an existing electric field. – Rijul Gupta Jan 16 '14 at 10:57
• @rijulgupta But the object you introduced is not neutral at all - it consists of two distinct charges with opposite signs. At a large distance - and this is what you see from the first term of the multipole expansion, $\frac{Q}{r}$ - every electric potential looks like that of a point particle (with charge $Q$ equal to the sum of the charges of the constituents $q_i$), but if you come closer (so you can't disregard terms of order $r^{-3}$ anymore), you have to resolve the inner structure of the charge distribution. – Wojciech Morawiec Jan 16 '14 at 11:01
• So do you agree that if you have a parallel plate capacitor with $E$ field set up inside, and you insert a neutral object in it, it would change the field from $E$ ? – Rijul Gupta Jan 16 '14 at 11:13
• @rijulgupta If by "neutral" object you mean an object composed of a positive and a negative charge that are spatially separated (so not a real neutral object), then the field inside the capacitor will be changed. This is exactly what happens when you insert a dielectric into a capacitor. – Wojciech Morawiec Jan 16 '14 at 11:24
• No I do not mean that, I mean an object which has uniform distribution of charges and hence is neutral , even in casenof dielectric I mean to say if you insert uncharged dielectric in small portion then why the rest of the field is not altered as well as that within the dielectric. – Rijul Gupta Jan 16 '14 at 11:26

The right answer is this: when a charged point is placed in the electric field that is generated by other charged objects other than the charged particle in question. Then we can use the formula $F=qE$ to calculate the force acted on the charged point. But not all charged particles with finite size can be approximated as a "charged point."