3. Find the total charge within a cube of side 1m centered at (0.5,0.5,0.5) . Assume that the electric field intensity within the cube is given by the expression overrightarrow (E)=E_(0)(1-e^-vert alpha yvert )overrightarrow (u) where E_(0)=2(mu V/m) and alpha =1(m^-1)

To find the total charge within the cube, we need to integrate the electric field intensity over the volume of the cube and then divide by the electric field constant (ε₀).<br /><br />The total charge (Q) can be calculated using the formula:<br /><br />\[ Q = \frac{1}{\varepsilon_0} \int_V \overrightarrow{E} \cdot d\overrightarrow{A} \]<br /><br />where \( \overrightarrow{E} \) is the electric field intensity, \( d\overrightarrow{A} \) is the differential area element, and \( \varepsilon_0 \) is the electric field constant.<br /><br />Given the electric field intensity expression:<br /><br />\[ \overrightarrow{E} = E_0 (1 - e^{-|\alpha y|}) \overrightarrow{u} \]<br /><br />where \( E_0 = 2 \, \mu V/m \) and \( \alpha = 1 \, m^{-1} \), we can substitute this into the integral:<br /><br />\[ Q = \frac{1}{\varepsilon_0} \int_V E_0 (1 - e^{-|\alpha y|}) \overrightarrow{u} \cdot d\overrightarrow{A} \]<br /><br />Since the electric field is uniform in the x and z directions and only varies in the y direction, we can simplify the integral to:<br /><br />\[ Q = \frac{E_0}{\varepsilon_0} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (1 - e^{-|\alpha y|}) \, dy \, dx \, dz \]<br /><br />Evaluating this triple integral will give us the total charge within the cube.