Two charges of 5mathrmQ$5\mathrm{Q}$ and -2mathrmQ$-2\mathrm{Q}$ are situated at the points (3a, 0)$(3a, 0)$ and (-5a, 0)$(-5a, 0)$ respectively. The electric flux through a sphere of radius '4a' having center at origin is:

### Related Formula Gauss's Law states: Phi = oint vecE cdot dvecA = fracq_textenclosedvarepsilon_0$$\Phi = \oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$$ ### Core Logic We have a sphere of radius R = 4a$R = 4a$ centered at (0,0)$(0,0)$. * Charge 5Q$5Q$ is located at (3a, 0)$(3a, 0)$. Since the distance from origin is 3a lt 4a$3a \lt 4a$, this charge lies **inside** the sphere. * Charge -2Q$-2Q$ is located at (-5a, 0)$(-5a, 0)$. Since the distance from origin is 5a gt 4a$5a \gt 4a$, this charge lies **outside** the sphere.

### Step 1: Calculate Enclosed Charge The net enclosed charge q_textenclosed$q_{\text{enclosed}}$ within the spherical boundary is: q_textenclosed = 5Q$$q_{\text{enclosed}} = 5Q$$ ### Step 2: Apply Gauss Law Using Gauss's Law, the total electric flux is: Phi = fracq_textenclosedvarepsilon_0 = frac5Qvarepsilon_0$$\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{5Q}{\varepsilon_0}$$ Therefore, the electric flux is frac5Qvarepsilon_0$\frac{5Q}{\varepsilon_0}$. ### Pattern Recognition Flux depends purely on charges situated *inside* the closed surface. Charges outside the Gaussian surface contribute absolutely zero net flux because every field line entering must also exit. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electric Charges and Fields