Question

a point charge is placed in the interior of a large insulating block with dielectric constant 2.0. the electric field produced by the point charge at a point 3 cm from the charge is 400 n/c. the magnitude of the electric field produced at that point by the polarization charge is
o 600 n/c
o 400 n/c
o 200 n/c
o 800 n/c
o 0 n/c
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Explanation:

Step1: Recall the relationship between electric - field in a dielectric and in vacuum

The electric field in a dielectric $E$ and the electric field in vacuum $E_0$ are related by $E=\frac{E_0}{\kappa}$, where $\kappa$ is the dielectric constant. Let the electric field produced by the point - charge in vacuum be $E_0$ and the electric field in the dielectric be $E$. The electric field due to polarization $E_p$ is related to $E$ and $E_0$ by $E = E_0 - E_p$.
We know that $E=\frac{E_0}{\kappa}$. Rearranging for $E_p$, we get $E_p=E_0 - E$. Since $E=\frac{E_0}{\kappa}$, then $E_p = E_0(1-\frac{1}{\kappa})$.

Step2: Identify the given values

We are given that $E_0 = 400\ N/C$ and $\kappa = 2.0$.

Step3: Calculate the electric field due to polarization

Substitute the values into the formula $E_p = E_0(1 - \frac{1}{\kappa})$.
$E_p=400(1-\frac{1}{2})$
$E_p = 400\times\frac{1}{2}=200\ N/C$

Answer:

200 N/C