A parallet plate capacitor of area of 2.0 mathrm(~m)^2 and relative permituity of 7 is charged to a pd of 240 mathrm(~V) . Calculate the capacitance and energy stored in the capacitor of plate ceparation (d) of 0.0004(C=30.9 mathrm(MF), E=0.809 mathrm(~g))

To calculate the capacitance of a parallel plate capacitor, we can use the formula:<br /><br />\[ C = \frac{\varepsilon \cdot A}{d} \]<br /><br />where:<br />- \( C \) is the capacitance,<br />- \( \varepsilon \) is the relative permittivity,<br />- \( A \) is the area of the plates, and<br />- \( d \) is the separation between the plates.<br /><br />Given that the area of the plates is \( 2.0 \, \text{m}^2 \), the relative permittivity is 7, and the separation between the plates is \( 0.0004 \, \text{m} \), we can substitute these values into the formula:<br /><br />\[ C = \frac{7 \cdot 2.0}{0.0004} \]<br /><br />\[ C = \frac{14}{0.0004} \]<br /><br />\[ C = 35000 \, \text{F} \]<br /><br />So, the capacitance of the capacitor is \( 35000 \, \text{F} \) or \( 35 \, \text{mF} \).<br /><br />To calculate the energy stored in the capacitor, we can use the formula:<br /><br />\[ E = \frac{1}{2} C V^2 \]<br /><br />where:<br />- \( E \) is the energy stored,<br />- \( C \) is the capacitance, and<br />- \( V \) is the voltage across the capacitor.<br /><br />Given that the capacitance is \( 35 \, \text{mF} \) and the voltage across the capacitor is \( 240 \, \text{V} \), we can substitute these values into the formula:<br /><br />\[ E = \frac{1}{2} \cdot 35 \cdot 240^2 \]<br /><br />\[ E = \frac{1}{2} \cdot 35 \cdot 57600 \]<br /><br />\[ E = 1008000 \, \text{J} \]<br /><br />So, the energy stored in the capacitor is \( 1008000 \, \text{J} \) or \( 1.008 \, \text{GJ} \).